,transcribers note3 ,many of the puzzles in this book assume a familiarity with the currency of ,great ,britain in the early #aijjs4 ,as this is likely not common knowledge for those outside ,britain 7and possibly many within17 ,i am including a chart of relative values4 ,the most common units used were3 the ,penny1 abbreviated3 d4 7from the ,roman penny1 denarius7 the ,shilling1 abbreviated3 s4 the ,pound1 abbreviated3 ,there was #ab ,pennies to a ,shilling and #bj ,shillings to a ,pound1 so there was #bdj ,pennies in a ,pound4 ,to further complicate things1 there were many coins which were various fractional values of ,pennies1 ,shillings or ,pounds4 ,farthing d4 ,half-penny d4 ,penny #a;d4 ,three-penny #c;d4 ,sixpence 7or tanner7 #f;d4 ,shilling 7or bob7 #as4 ,florin or two shilling piece #bs4 ,half-crown 7or half-dollar7 #bs4 #f;d4 ,double-florin #ds4 ,crown 7or dollar7 #es4 ,half-,sovereign #ajs4 ,sovereign 7or ,pound7 #a or #bjs4 ,this is by no means a comprehensive list1 but it should be adequate to solve the puzzles in this book4 ,exponents are represented in this text by 1 e4g4 '#c squared' is #cb4 ,numbers with fractional components 7other than 1 and 7 have a symbol separating the whole number component from the fraction4 ,it makes the fraction look odd1 but yeilds correct solutions no matter how it is interpreted4 ,e4,g41 #d and eleven twenty-thirds is #daa_/bc1 not #daa_/bc or #d-aa_/bc4 ,,amusements ,,in ,,mathematics by ,,henry ,,ernest ,,dudeney ,in ,mathematicks he was greater ,than ,tycho ,brahe or ,erra ,pater3 ,for he1 by geometrick scale1 ,could take the size of pots of ale2 ,resolve1 by sines and tangents1 straight1 ,if bread or butter wanted weight2 ,and wisely tell what hour o' th' day ,the clock does strike by algebra4 ,,butler',s ,hudibras4 #aiag ,,preface ,in issuing this volume of my ,mathematical ,puzzles1 of which some have appeared in periodicals and others are given here for the first time1 ,i must acknowledge the encouragement that ,i have received from many unknown correspondents1 at home and abroad1 who have expressed a desire to have the problems in a collected form1 with some of the solutions given at greater length than is possible in magazines and newspapers4 ,though ,i have included a few old puzzles that have interested the world for generations1 where ,i felt that there was something new to be said about them1 the problems are in the main original4 ,it is true that some of these have become widely known through the press1 and it is possible that the reader may be glad to know their source4 ,on the question of ,mathematical ,puzzles in general there is1 perhaps1 little more to be said than ,i have written elsewhere4 ,the history of the subject entails nothing short of the actual story of the beginnings and development of exact thinking in man4 ,the historian must start from the time when man first succeeded in counting his ten fingers and in dividing an apple into two approximately equal parts4 ,every puzzle that is worthy of consideration can be referred to mathematics and logic4 ,every man1 woman1 and child who tries to reason out the answer to the simplest puzzle is working1 though not of necessity consciously1 on mathematical lines4 ,even those puzzles that we have no way of attacking except by haphazard attempts can be brought under a method of what has been called glorified trial--a system of shortening our labours by avoiding or eliminating what our reason tells us is useless4 ,it is1 in fact1 not easy to say sometimes where the empirical begins and where it ends4 ,when a man says1 ,i have never solved a puzzle in my life1 it is difficult to know exactly what he means1 for every intelligent individual is doing it every day4 ,the unfortunate inmates of our lunatic asylums are sent there expressly because they cannot solve puzzles--because they have lost their powers of reason4 ,if there were no puzzles to solve1 there would be no questions to ask2 and if there were no questions to be asked1 what a world it would be6 ,we should all be equally omniscient1 and conversation would be useless and idle4 ,it is possible that some few exceedingly sober-minded mathematicians1 who are impatient of any terminology in their favourite science but the academic1 and who object to the elusive x and y appearing under any other names1 will have wished that various problems had been presented in a less popular dress and introduced with a less flippant phraseology4 ,i can only refer them to the first word of my title and remind them that we are primarily out to be amused--not1 it is true1 without some hope of picking up morsels of knowledge by the way4 ,if the manner is light1 ,i can only say1 in the words of ,touchstone1 that it is an ill-favoured thing1 sir1 but my own2 a poor humour of mine1 sir4 ,as for the question of difficulty1 some of the puzzles1 especially in the ,arithmetical and ,algebraical category1 are quite easy4 ,yet some of those examples that look the simplest should not be passed over without a little consideration1 for now and again it will be found that there is some more or less subtle pitfall or trap into which the reader may be apt to fall4 ,it is good exercise to cultivate the habit of being very wary over the exact wording of a puzzle4 ,it teaches exactitude and caution4 ,but some of the problems are very hard nuts indeed1 and not unworthy of the attention of the advanced mathematician4 ,readers will doubtless select according to their individual tastes4 ,in many cases only the mere answers are given4 ,this leaves the beginner something to do on his own behalf in working out the method of solution1 and saves space that would be wasted from the point of view of the advanced student4 ,on the other hand1 in particular cases where it seemed likely to interest1 ,i have given rather extensive solutions and treated problems in a general manner4 ,it will often be found that the notes on one problem will serve to elucidate a good many others in the book2 so that the reader's difficulties will sometimes be found cleared up as he advances4 ,where it is possible to say a thing in a manner that may be understanded of the people generally1 ,i prefer to use this simple phraseology1 and so engage the attention and interest of a larger public4 ,the mathematician will in such cases have no difficulty in expressing the matter under consideration in terms of his familiar symbols4 ,i have taken the greatest care in reading the proofs1 and trust that any errors that may have crept in are very few4 ,if any such should occur1 ,i can only plead1 in the words of ,horace1 that good ,homer sometimes nods1 or1 as the bishop put it1 ,not even the youngest curate in my diocese is infallible4 ,i have to express my thanks in particular to the proprietors of ,the ,strand ,magazine1 ,cassell's ,magazine1 ,the ,queen1 ,tit-,bits1 and ,the ,weekly ,dispatch for their courtesy in allowing me to reprint some of the puzzles that have appeared in their pages4 ,,the ,,authors' ,,club ,march #be1 #aiag ,,contents ,,preface v ,,arithmetical ,,and ,,algebraical ,,problems #a ,money ,puzzles #a ,age and ,kinship ,puzzles #f ,clock ,puzzles #i ,locomotion and ,speed ,puzzles #aa ,digital ,puzzles #ac ,various ,arithmetical and ,algebraical ,problems #ag ,,geometrical ,,problems #bg ,dissection ,puzzles #bg ,greek ,cross ,puzzles #bh ,various ,dissection ,puzzles #ce ,patchwork ,puzzles #df ,various ,geometrical ,puzzles #di ,,points ,,and ,,lines ,,problems #ef ,,moving ,,counter ,,problems #eh ,,unicursal ,,and ,,route ,,problems #fh ,,combination ,,and ,,group ,,problems #gf ,,chessboard ,,problems #he ,the ,chessboard #he ,statical ,chess ,puzzles #hh ,the ,guarded ,chessboard #ie ,dynamical ,chess ,puzzles #if ,various ,chess ,puzzles #aje ,,measuring1 ,,weighing1 ,,and ,,packing ,,puzzles #aji ,,crossing ,,river ,,problems #aab ,,problems ,,concerning ,,games #aad ,,puzzle ,,games #aag ,,magic ,,square ,,problems #aai ,subtracting1 ,multiplying1 and ,dividing ,magics #abd ,magic ,squares of ,primes #abe ,,mazes ,,and ,,how ,,to ,,thread ,,them #abg ,,the ,,paradox ,,party #acg ,,unclassified ,,problems #adb ,,solutions #adh ,,index #bec ,,amusements ,,in ,,mathematics4 ,,arithmetical ,,and ,,algebraical ,,problems4 ,and what was he8 ,forsooth1 a great arithmetician4 ,othello1 ,i4 i4 ,the puzzles in this department are roughly thrown together in classes for the convenience of the reader4 ,some are very easy1 others quite difficult4 ,but they are not arranged in any order of difficulty--and this is intentional1 for it is well that the solver should not be warned that a puzzle is just what it seems to be4 ,it may1 therefore1 prove to be quite as simple as it looks1 or it may contain some pitfall into which1 through want of care or over-confidence1 we may stumble4 ,also1 the arithmetical and algebraical puzzles are not separated in the manner adopted by some authors1 who arbitrarily require certain problems to be solved by one method or the other4 ,the reader is left to make his own choice and determine which puzzles are capable of being solved by him on purely arithmetical lines4 ,,money ,,puzzles4 ,put not your trust in money1 but put your money in trust4 ,,oliver ,,wendell ,,holmes4 #a4--;,a ,,post-,,office ,,perplexity4 ,in every business of life we are occasionally perplexed by some chance question that for the moment staggers us4 ,i quite pitied a young lady in a branch post-office when a gentleman entered and deposited a crown on the counter with this request3 ,please give me some twopenny stamps1 six times as many penny stamps1 and make up the rest of the money in twopence-halfpenny stamps4 ,for a moment she seemed bewildered1 then her brain cleared1 and with a smile she handed over stamps in exact fulfilment of the order4 ,how long would it have taken you to think it out8 #b4--,,youthful ,,precocity4 ,the precocity of some youths is surprising4 ,one is disposed to say on occasion1 ,that boy of yours is a genius1 and he is certain to do great things when he grows up2 but past experience has taught us that he invariably becomes quite an ordinary citizen4 ,it is so often the case1 on the contrary1 that the dull boy becomes a great man4 ,you never can tell4 ,nature loves to present to us these queer paradoxes4 ,it is well known that those wonderful lightning calculators1 who now and again surprise the world by their feats1 lose all their mysterious powers directly they are taught the elementary rules of arithmetic4 ,a boy who was demolishing a choice banana was approached by a young friend1 who1 regarding him with envious eyes1 asked1 ,how much did you pay for that banana1 ,fred8 ,the prompt answer was quite remarkable in its way3 ,the man what ,i bought it of receives just half as many sixpences for sixteen dozen dozen bananas as he gives bananas for a fiver4 ,now1 how long will it take the reader to say correctly just how much ,fred paid for his rare and refreshing fruit8 #c4--;,,at ,a ,,cattle ,,market4 ,three countrymen met at a cattle market4 ,look here1 said ,hodge to ,jakes1 ,i'll give you six of my pigs for one of your horses1 and then you'll have twice as many animals here as ,i've got4 ,if that's your way of doing business1 said ,durrant to ,hodge1 ,i'll give you fourteen of my sheep for a horse1 and then you'll have three times as many animals as ,i4 ,well1 ,i'll go better than that1 said ,jakes to ,durrant2 ,i'll give you four cows for a horse1 and then you'll have six times as many animals as ,i've got here4 ,no doubt this was a very primitive way of bartering animals1 but it is an interesting little puzzle to discover just how many animals ,jakes1 ,hodge1 and ,durrant must have taken to the cattle market4 #d4--,,the ,,beanfeast ,,puzzle4 ,a number of men went out together on a bean-feast4 ,there were four parties invited--namely1 #be cobblers1 #bj tailors1 #ah hatters1 and #ab glovers4 ,they spent altogether #f1 #acs4 ,it was found that five cobblers spent as much as four tailors2 that twelve tailors spent as much as nine hatters2 and that six hatters spent as much as eight glovers4 ,the puzzle is to find out how much each of the four parties spent4 #e4--;,a ,,queer ,,coincidence4 ,seven men1 whose names were ,adams1 ,baker1 ,carter1 ,dobson1 ,edwards1 ,francis1 and ,gudgeon1 were recently engaged in play4 ,the name of the particular game is of no consequence4 ,they had agreed that whenever a player won a game he should double the money of each of the other players--that is1 he was to give the players just as much money as they had already in their pockets4 ,they played seven games1 and1 strange to say1 each won a game in turn1 in the order in which their names are given4 ,but a more curious coincidence is this--that when they had finished play each of the seven men had exactly the same amount--two shillings and eightpence--in his pocket4 ,the puzzle is to find out how much money each man had with him before he sat down to play4 #f4--;,a ,,charitable ,,bequest4 ,a man left instructions to his executors to distribute once a year exactly fifty-five shillings among the poor of his parish2 but they were only to continue the gift so long as they could make it in different ways1 always giving eighteenpence each to a number of women and half a crown each to men4 ,during how many years could the charity be administered8 ,of course1 by different ways is meant a different number of men and women every time4 #g4--,,the ,,widow',s ,,legacy4 ,a gentleman who recently died left the sum of #h1jjj to be divided among his widow1 five sons1 and four daughters4 ,he directed that every son should receive three times as much as a daughter1 and that every daughter should have twice as much as their mother4 ,what was the widow's share8 #h4--;,,indiscriminate ,,charity4 ,a charitable gentleman1 on his way home one night1 was appealed to by three needy persons in succession for assistance4 ,to the first person he gave one penny more than half the money he had in his pocket2 to the second person he gave twopence more than half the money he then had in his pocket2 and to the third person he handed over threepence more than half of what he had left4 ,on entering his house he had only one penny in his pocket4 ,now1 can you say exactly how much money that gentleman had on him when he started for home8 #i4--,,the ,,two ,,aeroplanes4 ,a man recently bought two aeroplanes1 but afterwards found that they would not answer the purpose for which he wanted them4 ,so he sold them for #fjj each1 making a loss of #bj per cent4 on one machine and a profit of #bj per cent4 on the other4 ,did he make a profit on the whole transaction1 or a loss8 ,and how much8 #aj4--;,,buying ,,presents4 ,whom do you think ,i met in town last week1 ,brother ,william8 said ,uncle ,benjamin4 ,that old skinflint ,jorkins4 ,his family had been taking him around buying ,christmas presents4 ,he said to me1 ',why cannot the government abolish ,christmas1 and make the giving of presents punishable by law8 ,i came out this morning with a certain amount of money in my pocket1 and ,i find ,i have spent just half of it4 ,in fact1 if you will believe me1 ,i take home just as many shillings as ,i had pounds1 and half as many pounds as ,i had shillings4 ,it is monstrous6' ,can you say exactly how much money ,jorkins had spent on those presents8 #aa4--,,the ,,cyclists' ,,feast4 ',twas last ,bank ,holiday1 so ,i've been told1 ,some cyclists rode abroad in glorious weather4 ,resting at noon within a tavern old1 ,they all agreed to have a feast together4 ,put it all in one bill1 mine host1 they said1 ,for every man an equal share will pay4 ,the bill was promptly on the table laid1 ,and four pounds was the reckoning that day4 ,but1 sad to state1 when they prepared to square1 ',twas found that two had sneaked outside and fled4 ,so1 for two shillings more than his due share ,each honest man who had remained was bled4 ,they settled later with those rogues1 no doubt4 ,how many were they when they first set out8 #ab4--;,a ,,queer ,,thing ,,in ,,money4 ,it will be found that #ff1 #fs4 #f;d4 equals #ae1iah pence4 ,now1 the four #f's added together make #bd1 and the figures in #ae1iah also add to #bd4 ,it is a curious fact that there is only one other sum of money1 in pounds1 shillings1 and pence 7all similarly repetitions of one figure71 of which the digits shall add up the same as the digits of the amount in pence4 ,what is the other sum of money8 #ac4--;,a ,,new ,,money ,,puzzle4 ,the largest sum of money that can be written in pounds1 shillings1 pence1 and farthings1 using each of the nine digits once and only once1 is #ih1gfe1 #ds4 #c;d4 ,now1 try to discover the smallest sum of money that can be written down under precisely the same conditions4 ,there must be some value given for each denomination--pounds1 shillings1 pence1 and farthings--and the nought may not be used4 ,it requires just a little judgment and thought4 #ad4--,,square ,,money4 ,this is queer1 said ,mc,crank to his friend4 ,twopence added to twopence is fourpence1 and twopence multiplied by twopence is also fourpence4 ,of course1 he was wrong in thinking you can multiply money by money4 ,the multiplier must be regarded as an abstract number4 ,it is true that two feet multiplied by two feet will make four square feet4 ,similarly1 two pence multiplied by two pence will produce four square pence6 ,and it will perplex the reader to say what a square penny is4 ,but we will assume for the purposes of our puzzle that twopence multiplied by twopence is fourpence4 ,now1 what two amounts of money will produce the next smallest possible result1 the same in both cases1 when added or multiplied in this manner8 ,the two amounts need not be alike1 but they must be those that can be paid in current coins of the realm4 #ae4--,,pocket ,,money4 ,what is the largest sum of money--all in current silver coins and no four-shilling piece--that ,i could have in my pocket without being able to give change for a half-sovereign8 #af4--,,the ,,millionaire',s ,,perplexity4 ,mr4 ,morgan ,g4 ,bloomgarten1 the millionaire1 known in the ,states as the ,clam ,king1 had1 for his sins1 more money than he knew what to do with4 ,it bored him4 ,so he determined to persecute some of his poor but happy friends with it4 ,they had never done him any harm1 but he resolved to inoculate them with the source of all evil4 ,he therefore proposed to distribute a million dollars among them and watch them go rapidly to the bad4 ,but he was a man of strange fancies and superstitions1 and it was an inviolable rule with him never to make a gift that was not either one dollar or some power of seven--such as #g1 #di1 #cdc1 #b1dja1 which numbers of dollars are produced by simply multiplying sevens together4 ,another rule of his was that he would never give more than six persons exactly the same sum4 ,now1 how was he to distribute the #a1jjj1jjj dollars8 ,you may distribute the money among as many people as you like1 under the conditions given4 #ag4--,,the ,,puzzling ,,money-,,boxes4 ,four brothers--named ,john1 ,william1 ,charles1 and ,thomas--had each a money-box4 ,the boxes were all given to them on the same day1 and they at once put what money they had into them2 only1 as the boxes were not very large1 they first changed the money into as few coins as possible4 ,after they had done this1 they told one another how much money they had saved1 and it was found that if ,john had had #bs4 more in his box than at present1 if ,william had had #bs4 less1 if ,charles had had twice as much1 and if ,thomas had had half as much1 they would all have had exactly the same amount4 ,now1 when ,i add that all four boxes together contained #des41 and that there were only six coins in all in them1 it becomes an entertaining puzzle to discover just what coins were in each box4 #ah4--,,the ,,market ,,women4 ,a number of market women sold their various products at a certain price per pound 7different in every case71 and each received the same amount--#bs4 #b;d4 ,what is the greatest number of women there could have been8 ,the price per pound in every case must be such as could be paid in current money4 #ai4--,,the ,,new ,,year',s ,,eve ,,suppers4 ,the proprietor of a small ,london caf has given me some interesting figures4 ,he says that the ladies who come alone to his place for refreshment spend each on an average eighteenpence1 that the unaccompanied men spend half a crown each1 and that when a gentleman brings in a lady he spends half a guinea4 ,on ,new ,year's ,eve he supplied suppers to twenty-five persons1 and took five pounds in all4 ,now1 assuming his averages to have held good in every case1 how was his company made up on that occasion8 ,of course1 only single gentlemen1 single ladies1 and pairs 7a lady and gentleman7 can be supposed to have been present1 as we are not considering larger parties4 #bj4--;,,beef ,,and ,,sausages4 ,a neighbour of mine1 said ,aunt ,jane1 bought a certain quantity of beef at two shillings a pound1 and the same quantity of sausages at eighteenpence a pound4 ,i pointed out to her that if she had divided the same money equally between beef and sausages she would have gained two pounds in the total weight4 ,can you tell me exactly how much she spent8 ,of course1 it is no business of mine1 said ,mrs4 ,sunniborne2 but a lady who could pay such prices must be somewhat inexperienced in domestic economy4 ,i quite agree1 my dear1 ,aunt ,jane replied1 but you see that is not the precise point under discussion1 any more than the name and morals of the tradesman4 #ba4--;,a ,,deal ,,in ,,apples4 ,i paid a man a shilling for some apples1 but they were so small that ,i made him throw in two extra apples4 ,i find that made them cost just a penny a dozen less than the first price he asked4 ,how many apples did ,i get for my shilling8 #bb4--;,a ,,deal ,,in ,,eggs4 ,a man went recently into a dairyman's shop to buy eggs4 ,he wanted them of various qualities4 ,the salesman had new-laid eggs at the high price of fivepence each1 fresh eggs at one penny each1 eggs at a halfpenny each1 and eggs for electioneering purposes at a greatly reduced figure1 but as there was no election on at the time the buyer had no use for the last4 ,however1 he bought some of each of the three other kinds and obtained exactly one hundred eggs for eight and fourpence4 ,now1 as he brought away exactly the same number of eggs of two of the three qualities1 it is an interesting puzzle to determine just how many he bought at each price4 #bc4--,,the ,,christmas-,,boxes4 ,some years ago a man told me he had spent one hundred ,english silver coins in ,christmas-boxes1 giving every person the same amount1 and it cost him exactly #a1 #ajs4 #a;d4 ,can you tell just how many persons received the present1 and how he could have managed the distribution8 ,that odd penny looks queer1 but it is all right4 #bd4--;,a ,,shopping ,,perplexity4 ,two ladies went into a shop where1 through some curious eccentricity1 no change was given1 and made purchases amounting together to less than five shillings4 ,do you know1 said one lady1 ,i find ,i shall require no fewer than six current coins of the realm to pay for what ,i have bought4 ,the other lady considered a moment1 and then exclaimed3 ,by a peculiar coincidence1 ,i am exactly in the same dilemma4 ,then we will pay the two bills together4 ,but1 to their astonishment1 they still required six coins4 ,what is the smallest possible amount of their purchases--both different8 #be4--;,,chinese ,,money4 ,the ,chinese are a curious people1 and have strange inverted ways of doing things4 ,it is said that they use a saw with an upward pressure instead of a downward one1 that they plane a deal board by pulling the tool toward them instead of pushing it1 and that in building a house they first construct the roof and1 having raised that into position1 proceed to work downwards4 ,in money the currency of the country consists of taels of fluctuating value4 ,the tael became thinner and thinner until #b1jjj of them piled together made less than three inches in height4 ,the common cash consists of brass coins of varying thicknesses1 with a round1 square1 or triangular hole in the centre1 as in our illustration4 ,illustration ,these are strung on wires like buttons4 ,supposing that eleven coins with round holes are worth fifteen ching-changs1 that eleven with square holes are worth sixteen ching-changs1 and that eleven with triangular holes are worth seventeen ching-changs1 how can a ,chinaman give me change for half a crown1 using no coins other than the three mentioned8 ,a ching-chang is worth exactly twopence and four-fifteenths of a ching-chang4 #bf4--,,the ,,junior ,,clerk',s ,,puzzle4 ,two youths1 bearing the pleasant names of ,moggs and ,snoggs1 were employed as junior clerks by a merchant in ,mincing ,lane4 ,they were both engaged at the same salary--that is1 commencing at the rate of #ej a year1 payable half-yearly4 ,moggs had a yearly rise of #aj1 and ,snoggs was offered the same1 only he asked1 for reasons that do not concern our puzzle1 that he might take his rise at #b1 #ajs4 half-yearly1 to which his employer 7not1 perhaps1 unnaturally67 had no objection4 ,now we come to the real point of the puzzle4 ,moggs put regularly into the ,post ,office ,savings ,bank a certain proportion of his salary1 while ,snoggs saved twice as great a proportion of his1 and at the end of five years they had together saved #bfh1 #aes4 ,how much had each saved8 ,the question of interest can be ignored4 #bg4--;,,giving ,,change4 ,every one is familiar with the difficulties that frequently arise over the giving of change1 and how the assistance of a third person with a few coins in his pocket will sometimes help us to set the matter right4 ,here is an example4 ,an ,englishman went into a shop in ,new ,york and bought goods at a cost of thirty-four cents4 ,the only money he had was a dollar1 a three-cent piece1 and a two-cent piece4 ,the tradesman had only a half-dollar and a quarter-dollar4 ,but another customer happened to be present1 and when asked to help produced two dimes1 a five-cent piece1 a two-cent piece1 and a one-cent piece4 ,how did the tradesman manage to give change8 ,for the benefit of those readers who are not familiar with the ,american coinage1 it is only necessary to say that a dollar is a hundred cents and a dime ten cents4 ,a puzzle of this kind should rarely cause any difficulty if attacked in a proper manner4 #bh4--;,,defective ,,observation4 ,our observation of little things is frequently defective1 and our memories very liable to lapse4 ,a certain judge recently remarked in a case that he had no recollection whatever of putting the wedding-ring on his wife's finger4 ,can you correctly answer these questions without having the coins in sight8 ,on which side of a penny is the date given8 ,some people are so unobservant that1 although they are handling the coin nearly every day of their lives1 they are at a loss to answer this simple question4 ,if ,i lay a penny flat on the table1 how many other pennies can ,i place around it1 every one also lying flat on the table1 so that they all touch the first one8 ,the geometrician will1 of course1 give the answer at once1 and not need to make any experiment4 ,he will also know that1 since all circles are similar1 the same answer will necessarily apply to any coin4 ,the next question is a most interesting one to ask a company1 each person writing down his answer on a slip of paper1 so that no one shall be helped by the answers of others4 ,what is the greatest number of three-penny-pieces that may be laid flat on the surface of a half-crown1 so that no piece lies on another or overlaps the surface of the half-crown8 ,it is amazing what a variety of different answers one gets to this question4 ,very few people will be found to give the correct number4 ,of course the answer must be given without looking at the coins4 #bi4--,,the ,,broken ,,coins4 ,a man had three coins--a sovereign1 a shilling1 and a penny--and he found that exactly the same fraction of each coin had been broken away4 ,now1 assuming that the original intrinsic value of these coins was the same as their nominal value--that is1 that the sovereign was worth a pound1 the shilling worth a shilling1 and the penny worth a penny--what proportion of each coin has been lost if the value of the three remaining fragments is exactly one pound8 #cj4--,,two ,,questions ,,in ,,probabilities4 ,there is perhaps no class of puzzle over which people so frequently blunder as that which involves what is called the theory of probabilities4 ,i will give two simple examples of the sort of puzzle ,i mean4 ,they are really quite easy1 and yet many persons are tripped up by them4 ,a friend recently produced five pennies and said to me3 ,in throwing these five pennies at the same time1 what are the chances that at least four of the coins will turn up either all heads or all tails8 ,his own solution was quite wrong1 but the correct answer ought not to be hard to discover4 ,another person got a wrong answer to the following little puzzle which ,i heard him propound3 ,a man placed three sovereigns and one shilling in a bag4 ,how much should be paid for permission to draw one coin from it8 ,it is1 of course1 understood that you are as likely to draw any one of the four coins as another4 #ca4--;,,domestic ,,economy4 ,young ,mrs4 ,perkins1 of ,putney1 writes to me as follows3 ,i should be very glad if you could give me the answer to a little sum that has been worrying me a good deal lately4 ,here it is3 ,we have only been married a short time1 and now1 at the end of two years from the time when we set up housekeeping1 my husband tells me that he finds we have spent a third of his yearly income in rent1 rates1 and taxes1 one-half in domestic expenses1 and one-ninth in other ways4 ,he has a balance of #aij remaining in the bank4 ,i know this last1 because he accidentally left out his pass-book the other day1 and ,i peeped into it4 ,don't you think that a husband ought to give his wife his entire confidence in his money matters8 ,well1 ,i do2 and--will you believe it8--he has never told me what his income really is1 and ,i want1 very naturally1 to find out4 ,can you tell me what it is from the figures ,i have given you8 ,yes2 the answer can certainly be given from the figures contained in ,mrs4 ,perkins's letter4 ,and my readers1 if not warned1 will be practically unanimous in declaring the income to be--something absurdly in excess of the correct answer6 #cb4--,,the ,,excursion ,,ticket ,,puzzle4 ,when the big flaming placards were exhibited at the little provincial railway station1 announcing that the ,great ---- ,company would run cheap excursion trains to ,london for the ,christmas holidays1 the inhabitants of ,mudley-cum-,turmits were in quite a flutter of excitement4 ,half an hour before the train came in the little booking office was crowded with country passengers1 all bent on visiting their friends in the great ,metropolis4 ,the booking clerk was unaccustomed to dealing with crowds of such a dimension1 and he told me afterwards1 while wiping his manly brow1 that what caused him so much trouble was the fact that these rustics paid their fares in such a lot of small money4 ,he said that he had enough farthings to supply a ,west ,end draper with change for a week1 and a sufficient number of threepenny pieces for the congregations of three parish churches4 ,that excursion fare1 said he1 is nineteen shillings and ninepence1 and ,i should like to know in just how many different ways it is possible for such an amount to be paid in the current coin of this realm4 ,here1 then1 is a puzzle3 ,in how many different ways may nineteen shillings and ninepence be paid in our current coin8 ,remember that the fourpenny-piece is not now current4 #cc4--,,puzzle ,,in ,,reversals4 ,most people know that if you take any sum of money in pounds1 shillings1 and pence1 in which the number of pounds 7less than #ab7 exceeds that of the pence1 reverse it 7calling the pounds pence and the pence pounds71 find the difference1 then reverse and add this difference1 the result is always #ab1 #ahs4 #aa;d4 ,but if we omit the condition1 less than #ab1 and allow nought to represent shillings or pence--7#a7 ,what is the lowest amount to which the rule will not apply8 7#b7 ,what is the highest amount to which it will apply8 ,of course1 when reversing such a sum as #ad1 #aes4 #c;d4 it may be written #c1 #afs4 #b;d41 which is the same as #c1 #aes4 #ad;d4 #cd4--,,the ,,grocer ,,and ,,draper4 ,a country grocer and draper had two rival assistants1 who prided themselves on their rapidity in serving customers4 ,the young man on the grocery side could weigh up two one-pound parcels of sugar per minute1 while the drapery assistant could cut three one-yard lengths of cloth in the same time4 ,their employer1 one slack day1 set them a race1 giving the grocer a barrel of sugar and telling him to weigh up forty-eight one-pound parcels of sugar ,while the draper divided a roll of forty-eight yards of cloth into yard pieces4 ,the two men were interrupted together by customers for nine minutes1 but the draper was disturbed seventeen times as long as the grocer4 ,what was the result of the race8 #ce4--;,,judkins',s ,,cattle4 ,hiram ,b4 ,judkins1 a cattle-dealer of ,texas1 had five droves of animals1 consisting of oxen1 pigs1 and sheep1 with the same number of animals in each drove4 ,one morning he sold all that he had to eight dealers4 ,each dealer bought the same number of animals1 paying seventeen dollars for each ox1 four dollars for each pig1 and two dollars for each sheep2 and ,hiram received in all three hundred and one dollars4 ,what is the greatest number of animals he could have had8 ,and how many would there be of each kind8 #cf4--;,,buying ,,apples4 ,as the purchase of apples in small quantities has always presented considerable difficulties1 ,i think it well to offer a few remarks on this subject4 ,we all know the story of the smart boy who1 on being told by the old woman that she was selling her apples at four for threepence1 said3 ,let me see6 ,four for threepence2 that's three for twopence1 two for a penny1 one for nothing--,i'll take one6 ,there are similar cases of perplexity4 ,for example1 a boy once picked up a penny apple from a stall1 but when he learnt that the woman's pears were the same price he exchanged it1 and was about to walk off4 ,stop6 said the woman4 ,you haven't paid me for the pear6 ,no1 said the boy1 of course not4 ,i gave you the apple for it4 ,but you didn't pay for the apple6 ,bless the woman6 ,you don't expect me to pay for the apple and the pear too6 ,and before the poor creature could get out of the tangle the boy had disappeared4 ,then1 again1 we have the case of the man who gave a boy sixpence and promised to repeat the gift as soon as the youngster had made it into ninepence4 ,five minutes later the boy returned4 ,i have made it into ninepence1 he said1 at the same time handing his benefactor threepence4 ,how do you make that out8 he was asked4 ,i bought threepennyworth of apples4 ,but that does not make it into ninepence6 ,i should rather think it did1 was the boy's reply4 ,the apple woman has threepence1 hasn't she8 ,very well1 ,i have threepennyworth of apples1 and ,i have just given you the other threepence4 ,what's that but ninepence8 ,i cite these cases just to show that the small boy really stands in need of a little instruction in the art of buying apples4 ,so ,i will give a simple poser dealing with this branch of commerce4 ,an old woman had apples of three sizes for sale--one a penny1 two a penny1 and three a penny4 ,of course two of the second size and three of the third size were respectively equal to one apple of the largest size4 ,now1 a gentleman who had an equal number of boys and girls gave his children sevenpence to be spent amongst them all on these apples4 ,the puzzle is to give each child an equal distribution of apples4 ,how was the sevenpence spent1 and how many children were there8 #cg4--;,,buying ,,chestnuts4 ,though the following little puzzle deals with the purchase of chestnuts1 it is not itself of the chestnut type4 ,it is quite new4 ,at first sight it has certainly the appearance of being of the nonsense puzzle character1 but it is all right when properly considered4 ,a man went to a shop to buy chestnuts4 ,he said he wanted a pennyworth1 and was given five chestnuts4 ,it is not enough2 ,i ought to have a sixth1 he remarked6 ,but if ,i give you one chestnut more4 the shopman replied1 you will have five too many4 ,now1 strange to say1 they were both right4 ,how many chestnuts should the buyer receive for half a crown8 #ch4--,,the ,,bicycle ,,thief4 ,here is a little tangle that is perpetually cropping up in various guises4 ,a cyclist bought a bicycle for #ae and gave in payment a cheque for #be4 ,the seller went to a neighbouring shopkeeper and got him to change the cheque for him1 and the cyclist1 having received his #aj change1 mounted the machine and disappeared4 ,the cheque proved to be valueless1 and the salesman was requested by his neighbour to refund the amount he had received4 ,to do this1 he was compelled to borrow the #be from a friend1 as the cyclist forgot to leave his address1 and could not be found4 ,now1 as the bicycle cost the salesman #aa1 how much money did he lose altogether8 #ci4--,,the ,,costermonger',s ,,puzzle4 ,how much did yer pay for them oranges1 ,bill8 ,i ain't a-goin' to tell yer1 ,jim4 ,but ,i beat the old cove down fourpence a hundred4 ,what good did that do yer8 ,well1 it meant five more oranges on every ten shillin's-worth4 ,now1 what price did ,bill actually pay for the oranges8 ,there is only one rate that will fit in with his statements4 ,,age ,,and ,,kinship ,,puzzles4 ,the days of our years are threescore years and ten4 --,psalm xc4 #aj4 ,for centuries it has been a favourite method of propounding arithmetical puzzles to pose them in the form of questions as to the age of an individual4 ,they generally lend themselves to very easy solution by the use of algebra1 though often the difficulty lies in stating them correctly4 ,they may be made very complex and may demand considerable ingenuity1 but no general laws can well be laid down for their solution4 ,the solver must use his own sagacity4 ,as for puzzles in relationship or kinship1 it is quite curious how bewildering many people find these things4 ,even in ordinary conversation1 some statement as to relationship1 which is quite clear in the mind of the speaker1 will immediately tie the brains of other people into knots4 ,such expressions as ,he is my uncle's son-in-law's sister convey absolutely nothing to some people without a detailed and laboured explanation4 ,in such cases the best course is to sketch a brief genealogical table1 when the eye comes immediately to the assistance of the brain4 ,in these days1 when we have a growing lack of respect for pedigrees1 most people have got out of the habit of rapidly drawing such tables1 which is to be regretted1 as they would save a lot of time and brain racking on occasions4 #dj4--,,mamma',s ,,age4 ,tommy3 ,how old are you1 mamma8 ,mamma3 ,let me think1 ,tommy4 ,well1 our three ages add up to exactly seventy years4 ,tommy3 ,that's a lot1 isn't it8 ,and how old are you1 papa8 ,papa3 ,just six times as old as you1 my son4 ,tommy3 ,shall ,i ever be half as old as you1 papa8 ,papa3 ,yes1 ,tommy2 and when that happens our three ages will add up to exactly twice as much as to-day4 ,tommy3 ,and supposing ,i was born before you1 papa2 and supposing mamma had forgot all about it1 and hadn't been at home when ,i came2 and supposing-- ,mamma3 ,supposing1 ,tommy1 we talk about bed4 ,come along1 darling4 ,you'll have a headache4 ,now1 if ,tommy had been some years older he might have calculated the exact ages of his parents from the information they had given him4 ,can you find out the exact age of mamma8 #da4--,,their ,,ages4 ,my husband's age1 remarked a lady the other day1 is represented by the figures of my own age reversed4 ,he is my senior1 and the difference between our ages is one-eleventh of their sum4 #db4--,,the ,,family ,,ages4 ,when the ,smileys recently received a visit from the favourite uncle1 the fond parents had all the five children brought into his presence4 ,first came ,billie and little ,gertrude1 and the uncle was informed that the boy was exactly twice as old as the girl4 ,then ,henrietta arrived1 and it was pointed out that the combined ages of herself and ,gertrude equalled twice the age of ,billie4 ,then ,charlie came running in1 and somebody remarked that now the combined ages of the two boys were exactly twice the combined ages of the two girls4 ,the uncle was expressing his astonishment at these coincidences when ,janet came in4 ,ah6 uncle1 she exclaimed1 you have actually arrived on my twenty-first birthday6 ,to this ,mr4 ,smiley added the final staggerer3 ,yes1 and now the combined ages of the three girls are exactly equal to twice the combined ages of the two boys4 ,can you give the age of each child8 #dc4--,,mrs4 ,,timpkins',s ,,age4 ,edwin3 ,do you know1 when the ,timpkinses married eighteen years ago ,timpkins was three times as old as his wife1 and to-day he is just twice as old as she8 ,angelina3 ,then how old was ,mrs4 ,timpkins on the wedding day8 ,can you answer ,angelina's question8 #dd--;,a ,,census ,,puzzle4 ,mr4 and ,mrs4 ,jorkins have fifteen children1 all born at intervals of one year and a half4 ,miss ,ada ,jorkins1 the eldest1 had an objection to state her age to the census man1 but she admitted that she was just seven times older than little ,johnnie1 the youngest of all4 ,what was ,ada's age8 ,do not too hastily assume that you have solved this little poser4 ,you may find that you have made a bad blunder6 #de4--,,mother ,,and ,,daughter4 ,mother1 ,i wish you would give me a bicycle1 said a girl of twelve the other day4 ,i do not think you are old enough yet1 my dear1 was the reply4 ,when ,i am only three times as old as you are you shall have one4 ,now1 the mother's age is forty-five years4 ,when may the young lady expect to receive her present8 #df4--,,mary ,,and ,,marmaduke4 ,marmaduke3 ,do you know1 dear1 that in seven years' time our combined ages will be sixty-three years8 ,mary3 ,is that really so8 ,and yet it is a fact that when you were my present age you were twice as old as ,i was then4 ,i worked it out last night4 ,now1 what are the ages of ,mary and ,marmaduke8 #dg--,,rover',s ,,age4 ,now1 then1 ,tommy1 how old is ,rover8 ,mildred's young man asked her brother4 ,well1 five years ago1 was the youngster's reply1 sister was four times older than the dog1 but now she is only three times as old4 ,can you tell ,rover's age8 #dh4--;,,concerning ,,tommy',s ,,age4 ,tommy ,smart was recently sent to a new school4 ,on the first day of his arrival the teacher asked him his age1 and this was his curious reply3 ,well1 you see1 it is like this4 ,at the time ,i was born--,i forget the year--my only sister1 ,ann1 happened to be just one-quarter the age of mother1 and she is now one-third the age of father4 ,that's all very well1 said the teacher1 but what ,i want is not the age of your sister ,ann1 but your own age4 ,i was just coming to that1 ,tommy answered2 ,i am just a quarter of mother's present age1 and in four years' time ,i shall be a quarter the age of father4 ,isn't that funny8 ,this was all the information that the teacher could get out of ,tommy ,smart4 ,could you have told1 from these facts1 what was his precise age8 ,it is certainly a little puzzling4 #di4--,,next-,,door ,,neighbours4 ,there were two families living next door to one another at ,tooting ,bec--the ,jupps and the ,simkins4 ,the united ages of the four ,jupps amounted to one hundred years1 and the united ages of the ,simkins also amounted to the same4 ,it was found in the case of each family that the sum obtained by adding the squares of each of the children's ages to the square of the mother's age equalled the square of the father's age4 ,in the case of the ,jupps1 however1 ,julia was one year older than her brother ,joe1 whereas ,sophy ,simkin was two years older than her brother ,sammy4 ,what was the age of each of the eight individuals8 #ej4--,,the ,,bag ,,of ,,nuts4 ,three boys were given a bag of nuts as a ,christmas present1 and it was agreed that they should be divided in proportion to their ages1 which together amounted to #ag years4 ,now the bag contained #ggj nuts1 and as often as ,herbert took four ,robert took three1 and as often as ,herbert took six ,christopher took seven4 ,the puzzle is to find out how many nuts each had1 and what were the boys' respective ages4 #ea4--;,,how ,,old ,,was ,,mary8 ,here is a funny little age problem1 by the late ,sam ,loyd1 which has been very popular in the ,united ,states4 ,can you unravel the mystery8 ,the combined ages of ,mary and ,ann are forty-four years1 and ,mary is twice as old as ,ann was when ,mary was half as old as ,ann will be when ,ann is three times as old as ,mary was when ,mary was three times as old as ,ann4 ,how old is ,mary8 ,that is all1 but can you work it out8 ,if not1 ask your friends to help you1 and watch the shadow of bewilderment creep over their faces as they attempt to grip the intricacies of the question4 #eb4--,,queer ,,relationships4 ,speaking of relationships1 said the ,parson at a certain dinner-party1 our legislators are getting the marriage law into a frightful tangle1 ,here1 for example1 is a puzzling case that has come under my notice4 ,two brothers married two sisters4 ,one man died and the other man's wife also died4 ,then the survivors married4 ,the man married his deceased wife's sister under the recent ,act8 put in the ,lawyer4 ,exactly4 ,and therefore1 under the civil law1 he is legally married and his child is legitimate4 ,but1 you see1 the man is the woman's deceased husband's brother1 and therefore1 also under the civil law1 she is not married to him and her child is illegitimate4 ,he is married to her and she is not married to him6 said the ,doctor4 ,quite so4 ,and the child is the legitimate son of his father1 but the illegitimate son of his mother4 ,undoubtedly 'the law is a hass1' the ,artist exclaimed1 if ,i may be permitted to say so1 he added1 with a bow to the ,lawyer4 ,certainly1 was the reply4 ,we lawyers try our best to break in the beast to the service of man4 ,our legislators are responsible for the breed4 ,and this reminds me1 went on the ,parson1 of a man in my parish who married the sister of his widow4 ,this man-- ,stop a moment1 sir1 said the ,professor4 ,married the sister of his widow8 ,do you marry dead men in your parish8 ,no2 but ,i will explain that later4 ,well1 this man has a sister of his own4 ,their names are ,stephen ,brown and ,jane ,brown4 ,last week a young fellow turned up whom ,stephen introduced to me as his nephew4 ,naturally1 ,i spoke of ,jane as his aunt1 but1 to my astonishment1 the youth corrected me1 assuring me that1 though he was the nephew of ,stephen1 he was not the nephew of ,jane1 the sister of ,stephen4 ,this perplexed me a good deal1 but it is quite correct4 ,the ,lawyer was the first to get at the heart of the mystery4 ,what was his solution8 #ec4--;,,heard ,,on ,,the ,,tube ,,railway4 ,first ,lady3 ,and was he related to you1 dear8 ,second ,lady3 ,oh1 yes4 ,you see1 that gentleman's mother was my mother's mother-in-law1 but he is not on speaking terms with my papa4 ,first ,lady3 ,oh1 indeed6 7,but you could see that she was not much wiser47 ,how was the gentleman related to the ,second ,lady8 #ed4--;,a ,,family ,,party4 ,a certain family party consisted of #a grandfather1 #a grandmother1 #b fathers1 #b mothers1 #d children1 #c grandchildren1 #a brother1 #b sisters1 #b sons1 #b daughters1 #a father-in-law1 #a mother-in-law1 and #a daughter-in-law4 ,twenty-three people1 you will say4 ,no2 there were only seven persons present4 ,can you show how this might be8 #ee4--;,a ,,mixed ,,pedigree4 ,joseph ,bloggs3 ,i can't follow it1 my dear boy4 ,it makes me dizzy6 ,john ,snoggs3 ,it's very simple4 ,listen again6 ,you happen to be my father's brother-in-law1 my brother's father-in-law1 and also my father-in-law's brother4 ,you see1 my father was-- ,but ,mr4 ,bloggs refused to hear any more4 ,can the reader show how this extraordinary triple relationship might have come about8 #ef4--,,wilson',s ,,poser4 ,speaking of perplexities-- said ,mr4 ,wilson1 throwing down a magazine on the table in the commercial room of the ,railway ,hotel4 ,who was speaking of perplexities8 inquired ,mr4 ,stubbs4 ,well1 then1 reading about them1 if you want to be exact--it just occurred to me that perhaps you three men may be interested in a little matter connected with myself4 ,it was ,christmas ,eve1 and the four commercial travellers were spending the holiday at ,grassminster4 ,probably each suspected that the others had no homes1 and perhaps each was conscious of the fact that he was in that predicament himself4 ,in any case they seemed to be perfectly comfortable1 and as they drew round the cheerful fire the conversation became general4 ,what is the difficulty8 asked ,mr4 ,packhurst4 ,there's no difficulty in the matter1 when you rightly understand it4 ,it is like this4 ,a man named ,parker had a flying-machine that would carry two4 ,he was a venturesome sort of chap--reckless1 ,i should call him--and he had some bother in finding a man willing to risk his life in making an ascent with him4 ,however1 an uncle of mine thought he would chance it1 and one fine morning he took his seat in the machine and she started off well4 ,when they were up about a thousand feet1 my nephew suddenly-- ,here1 stop1 ,wilson6 ,what was your nephew doing there8 ,you said your uncle1 interrupted ,mr4 ,stubbs4 ,did ,i8 ,well1 it does not matter4 ,my nephew suddenly turned to ,parker and said that the engine wasn't running well1 so ,parker called out to my uncle-- ,look here1 broke in ,mr4 ,waterson1 we are getting mixed4 ,was it your uncle or your nephew8 ,let's have it one way or the other4 ,what ,i said is quite right4 ,parker called out to my uncle to do something or other1 when my nephew-- ,there you are again1 ,wilson1 cried ,mr4 ,stubbs2 once for all1 are we to understand that both your uncle and your nephew were on the machine8 ,certainly4 ,i thought ,i made that clear4 ,where was ,i8 ,well1 my nephew shouted back to ,parker-- ,phew6 ,i'm sorry to interrupt you again1 ,wilson1 but we can't get on like this4 ,is it true that the machine would only carry two8 ,of course4 ,i said at the start that it only carried two4 ,then what in the name of aerostation do you mean by saying that there were three persons on board8 shouted ,mr4 ,stubbs4 ,who said there were three8 ,you have told us that ,parker1 your uncle1 and your nephew went up on this blessed flying-machine4 ,that's right4 ,and the thing would only carry two6 ,right again4 ,wilson1 ,i have known you for some time as a truthful man and a temperate man1 said ,mr4 ,stubbs1 solemnly4 ,but ,i am afraid since you took up that new line of goods you have overworked yourself4 ,half a minute1 ,stubbs1 interposed ,mr4 ,waterson4 ,i see clearly where we all slipped a cog4 ,of course1 ,wilson1 you meant us to understand that ,parker is either your uncle or your nephew4 ,now we shall be all right if you will just tell us whether ,parker is your uncle or nephew4 ,he is no relation to me whatever4 ,the three men sighed and looked anxiously at one another4 ,mr4 ,stubbs got up from his chair to reach the matches1 ,mr4 ,packhurst proceeded to wind up his watch1 and ,mr4 ,waterson took up the poker to attend to the fire4 ,it was an awkward moment1 for at the season of goodwill nobody wished to tell ,mr4 ,wilson exactly what was in his mind4 ,it's curious1 said ,mr4 ,wilson1 very deliberately1 and it's rather sad1 how thick-headed some people are4 ,you don't seem to grip the facts4 ,it never seems to have occurred to either of you that my uncle and my nephew are one and the same man4 ,what6 exclaimed all three together4 ,yes2 ,david ,george ,linklater is my uncle1 and he is also my nephew4 ,consequently1 ,i am both his uncle and nephew4 ,queer1 isn't it8 ,i'll explain how it comes about4 ,mr4 ,wilson put the case so very simply that the three men saw how it might happen without any marriage within the prohibited degrees4 ,perhaps the reader can work it out for himself4 ,,clock ,,puzzles4 ,look at the clock6 ,ingoldsby ,legends4 ,in considering a few puzzles concerning clocks and watches1 and the times recorded by their hands under given conditions1 it is well that a particular convention should always be kept in mind4 ,it is frequently the case that a solution requires the assumption that the hands can actually record a time involving a minute fraction of a second4 ,such a time1 of course1 cannot be really indicated4 ,is the puzzle1 therefore1 impossible of solution8 ,the conclusion deduced from a logical syllogism depends for its truth on the two premises assumed1 and it is the same in mathematics4 ,certain things are antecedently assumed1 and the answer depends entirely on the truth of those assumptions4 ,if two horses1 says ,lagrange1 can pull a load of a certain weight1 it is natural to suppose that four horses could pull a load of double that weight1 six horses a load of three times that weight4 ,yet1 strictly speaking1 such is not the case4 ,for the inference is based on the assumption that the four horses pull alike in amount and direction1 which in practice can scarcely ever be the case4 ,it so happens that we are frequently led in our reckonings to results which diverge widely from reality4 ,but the fault is not the fault of mathematics2 for mathematics always gives back to us exactly what we have put into it4 ,the ratio was constant according to that supposition4 ,the result is founded upon that supposition4 ,if the supposition is false the result is necessarily false4 ,if one man can reap a field in six days1 we say two men will reap it in three days1 and three men will do the work in two days4 ,we here assume1 as in the case of ,lagrange's horses1 that all the men are exactly equally capable of work4 ,but we assume even more than this4 ,for when three men get together they may waste time in gossip or play2 or1 on the other hand1 a spirit of rivalry may spur them on to greater diligence4 ,we may assume any conditions we like in a problem1 provided they be clearly expressed and understood1 and the answer will be in accordance with those conditions4 #eg4--,,what ,,was ,,the ,,time8 ,i say1 ,rackbrane1 what is the time8 an acquaintance asked our friend the professor the other day4 ,the answer was certainly curious4 ,if you add one quarter of the time from noon till now to half the time from now till noon to-morrow1 you will get the time exactly4 ,what was the time of day when the professor spoke8 #eh4--;,a ,,time ,,puzzle4 ,how many minutes is it until six o'clock if fifty minutes ago it was four times as many minutes past three o'clock8 #ei4--;,a ,,puzzling ,,watch4 ,a friend pulled out his watch and said1 ,this watch of mine does not keep perfect time2 ,i must have it seen to4 ,i have noticed that the minute hand and the hour hand are exactly together every sixty-five minutes4 ,does that watch gain or lose1 and how much per hour8 #fj4--,,the ,,wapshaw',s ,,wharf ,,mystery4 ,there was a great commotion in ,lower ,thames ,street on the morning of ,january #ab1 #ahhg4 ,when the early members of the staff arrived at ,wapshaw's ,wharf they found that the safe had been broken open1 a considerable sum of money removed1 and the offices left in great disorder4 ,the night watchman was nowhere to be found1 but nobody who had been acquainted with him for one moment suspected him to be guilty of the robbery4 ,in this belief the proprietors were confirmed when1 later in the day1 they were informed that the poor fellow's body had been picked up by the ,river ,police4 ,certain marks of violence pointed to the fact that he had been brutally attacked and thrown into the river4 ,a watch found in his pocket had stopped1 as is invariably the case in such circumstances1 and this was a valuable clue to the time of the outrage4 ,but a very stupid officer 7and we invariably find one or two stupid individuals in the most intelligent bodies of men7 had actually amused himself by turning the hands round and round1 trying to set the watch going again4 ,after he had been severely reprimanded for this serious indiscretion1 he was asked whether he could remember the time that was indicated by the watch when found4 ,he replied that he could not1 but he recollected that the hour hand and minute hand were exactly together1 one above the other1 and the second hand had just passed the forty-ninth second4 ,more than this he could not remember4 ,what was the exact time at which the watchman's watch stopped8 ,the watch is1 of course1 assumed to have been an accurate one4 #fa4--;,,changing ,,places4 ,illustration ,the above clock face indicates a little before #db minutes past #d4 ,the hands will again point at exactly the same spots a little after #bc minutes past #h4 ,in fact1 the hands will have changed places4 ,how many times do the hands of a clock change places between three o'clock p4m4 and midnight8 ,and out of all the pairs of times indicated by these changes1 what is the exact time when the minute hand will be nearest to the point ,,ix8 #fb4--,,the ,,club ,,clock4 ,one of the big clocks in the ,cogitators' ,club was found the other night to have stopped just when1 as will be seen in the illustration1 the second hand was exactly midway between the other two hands4 ,one of the members proposed to some of his friends that they should tell him the exact time when 7if the clock had not stopped7 the second hand would next again have been midway between the minute hand and the hour hand4 ,can you find the correct time that it would happen8 ,illustration #fc4--,,the ,,stop-,,watch4 ,illustration ,we have here a stop-watch with three hands4 ,the second hand1 which travels once round the face in a minute1 is the one with the little ring at its end near the centre4 ,our dial indicates the exact time when its owner stopped the watch4 ,you will notice that the three hands are nearly equidistant4 ,the hour and minute hands point to spots that are exactly a third of the circumference apart1 but the second hand is a little too advanced4 ,an exact equidistance for the three hands is not possible4 ,now1 we want to know what the time will be when the three hands are next at exactly the same distances as shown from one another4 ,can you state the time8 #fd4--,,the ,,three ,,clocks4 ,on ,friday1 ,april #a1 #ahih1 three new clocks were all set going precisely at the same time--twelve noon4 ,at noon on the following day it was found that clock ,a had kept perfect time1 that clock ,b had gained exactly one minute1 and that clock ,c had lost exactly one minute4 ,now1 supposing that the clocks ,b and ,c had not been regulated1 but all three allowed to go on as they had begun1 and that they maintained the same rates of progress without stopping1 on what date and at what time of day would all three pairs of hands again point at the same moment at twelve o'clock8 #fe4--,,the ,,railway ,,station ,,clock4 ,a clock hangs on the wall of a railway station1 #ga ft4 #i in4 long and #aj ft4 #d in4 high4 ,those are the dimensions of the wall1 not of the clock6 ,while waiting for a train we noticed that the hands of the clock were pointing in opposite directions1 and were parallel to one of the diagonals of the wall4 ,what was the exact time8 #ff4--,,the ,,village ,,simpleton4 ,a facetious individual who was taking a long walk in the country came upon a yokel sitting on a stile4 ,as the gentleman was not quite sure of his road1 he thought he would make inquiries of the local inhabitant2 but at the first glance he jumped too hastily to the conclusion that he had dropped on the village idiot4 ,he therefore decided to test the fellow's intelligence by first putting to him the simplest question he could think of1 which was1 ,what day of the week is this1 my good man8 ,the following is the smart answer that he received3-- ,when the day after to-morrow is yesterday1 to-day will be as far from ,sunday as to-day was from ,sunday when the day before yesterday was to-morrow4 ,can the reader say what day of the week it was8 ,it is pretty evident that the countryman was not such a fool as he looked4 ,the gentleman went on his road a puzzled but a wiser man4 ,,locomotion ,,and ,,speed ,,puzzles4 ,the race is not to the swift4--,ecclesiastes ix4 ,,ii4 #fg4--;,,average ,,speed4 ,in a recent motor ride it was found that we had gone at the rate of ten miles an hour1 but we did the return journey over the same route1 owing to the roads being more clear of traffic1 at fifteen miles an hour4 ,what was our average speed8 ,do not be too hasty in your answer to this simple little question1 or it is pretty certain that you will be wrong4 #fh4--,,the ,,two ,,trains4 ,i put this little question to a stationmaster1 and his correct answer was so prompt that ,i am convinced there is no necessity to seek talented railway officials in ,america or elsewhere4 ,two trains start at the same time1 one from ,london to ,liverpool1 the other from ,liverpool to ,london4 ,if they arrive at their destinations one hour and four hours respectively after passing one another1 how much faster is one train running than the other8 #fi4--,,the ,,three ,,villages4 ,i set out the other day to ride in a motor-car from ,acrefield to ,butterford1 but by mistake ,i took the road going via ,cheesebury1 which is nearer ,acrefield than ,butterford1 and is twelve miles to the left of the direct road ,i should have travelled4 ,after arriving at ,butterford ,i found that ,i had gone thirty-five miles4 ,what are the three distances between these villages1 each being a whole number of miles8 ,i may mention that the three roads are quite straight4 #gj4--;,,drawing ,,her ,,pension4 ,speaking of odd figures1 said a gentleman who occupies some post in a ,government office1 one of the queerest characters ,i know is an old lame widow who climbs up a hill every week to draw her pension at the village post office4 ,she crawls up at the rate of a mile and a half an hour and comes down at the rate of four and a half miles an hour1 so that it takes her just six hours to make the double journey4 ,can any of you tell me how far it is from the bottom of the hill to the top8 ,illustration #ga4--,,sir ,,edwyn ,,de ,,tudor4 ,in the illustration we have a sketch of ,sir ,edwyn de ,tudor going to rescue his lady-love1 the fair ,isabella1 who was held a captive by a neighbouring wicked baron4 ,sir ,edwyn calculated that if he rode fifteen miles an hour he would arrive at the castle an hour too soon1 while if he rode ten miles an hour he would get there just an hour too late4 ,now1 it was of the first importance that he should arrive at the exact time appointed1 in order that the rescue that he had planned should be a success1 and the time of the tryst was five o'clock1 when the captive lady would be taking her afternoon tea4 ,the puzzle is to discover exactly how far ,sir ,edwyn de ,tudor had to ride4 #gb4--,,the ,,hydroplane ,,question4 ,the inhabitants of ,slocomb-on-,sea were greatly excited over the visit of a certain flying man4 ,all the town turned out to see the flight of the wonderful hydroplane1 and1 of course1 ,dobson and his family were there4 ,master ,tommy was in good form1 and informed his father that ,englishmen made better airmen than ,scotsmen and ,irishmen because they are not so heavy4 ,how do you make that out8 asked ,mr4 ,dobson4 ,well1 you see1 ,tommy replied1 it is true that in ,ireland there are men of ,cork and in ,scotland men of ,ayr1 which is better still1 but in ,england there are lightermen4 ,unfortunately it had to be explained to ,mrs4 ,dobson1 and this took the edge off the thing4 ,the hydroplane flight was from ,slocomb to the neighbouring watering-place ,poodleville--five miles distant4 ,but there was a strong wind1 which so helped the airman that he made the outward journey in the short time of ten minutes1 though it took him an hour to get back to the starting point at ,slocomb1 with the wind dead against him4 ,now1 how long would the ten miles have taken him if there had been a perfect calm8 ,of course1 the hydroplane's engine worked uniformly throughout4 #gc4--;,,donkey ,,riding4 ,during a visit to the seaside ,tommy and ,evangeline insisted on having a donkey race over the mile course on the sands4 ,mr4 ,dobson and some of his friends whom he had met on the beach acted as judges1 but1 as the donkeys were familiar acquaintances and declined to part company the whole way1 a dead heat was unavoidable4 ,however1 the judges1 being stationed at different points on the course1 which was marked off in quarter-miles1 noted the following results3--,the first three-quarters were run in six and three-quarter minutes1 the first half-mile took the same time as the second half1 and the third quarter was run in exactly the same time as the last quarter4 ,from these results ,mr4 ,dobson amused himself in discovering just how long it took those two donkeys to run the whole mile4 ,can you give the answer8 #gd4--,,the ,,basket ,,of ,,potatoes4 ,a man had a basket containing fifty potatoes4 ,he proposed to his son1 as a little recreation1 that he should place these potatoes on the ground in a straight line4 ,the distance between the first and second potatoes was to be one yard1 between the second and third three yards1 between the third and fourth five yards1 between the fourth and fifth seven yards1 and so on--an increase of two yards for every successive potato laid down4 ,then the boy was to pick them up and put them in the basket one at a time1 the basket being placed beside the first potato4 ,how far would the boy have to travel to accomplish the feat of picking them all up8 ,we will not consider the journey involved in placing the potatoes1 so that he starts from the basket with them all laid out4 #ge4--,,the ,,passenger',s ,,fare4 ,at first sight you would hardly think there was matter for dispute in the question involved in the following little incident1 yet it took the two persons concerned some little time to come to an agreement4 ,mr4 ,smithers hired a motor-car to take him from ,addleford to ,clinkerville and back again for #c4 ,at ,bakenham1 just midway1 he picked up an acquaintance1 ,mr4 ,tompkins1 and agreed to take him on to ,clinkerville and bring him back to ,bakenham on the return journey4 ,how much should he have charged the passenger8 ,that is the question4 ,what was a reasonable fare for ,mr4 ,tompkins8 ,,digital ,,puzzles4 ,nine worthies were they called4 ,,dryden3 ,the ,flower and the ,leaf4 ,i give these puzzles1 dealing with the nine digits1 a class to themselves1 because ,i have always thought that they deserve more consideration than they usually receive4 ,beyond the mere trick of casting out nines1 very little seems to be generally known of the laws involved in these problems1 and yet an acquaintance with the properties of the digits often supplies1 among other uses1 a certain number of arithmetical checks that are of real value in the saving of labour4 ,let me give just one example--the first that occurs to me4 ,if the reader were required to determine whether or not #ae1gfc1ecj1afc1bhi is a square number1 how would he proceed8 ,if the number had ended with a #b1 #c1 #g1 or #h in the digits place1 of course he would know that it could not be a square1 but there is nothing in its apparent form to prevent its being one4 ,i suspect that in such a case he would set to work1 with a sigh or a groan1 at the laborious task of extracting the square root4 ,yet if he had given a little attention to the study of the digital properties of numbers1 he would settle the question in this simple way4 ,the sum of the digits is #ei1 the sum of which is #ad1 the sum of which is #e 7which ,i call the digital root71 and therefore ,i know that the number cannot be a square1 and for this reason4 ,the digital root of successive square numbers from #a upwards is always #a1 #d1 #g1 or #i1 and can never be anything else4 ,in fact1 the series1 #a1 #d1 #i1 #g1 #g1 #i1 #d1 #a1 #i1 is repeated into infinity4 ,the analogous series for triangular numbers is #a1 #c1 #f1 #a1 #f1 #c1 #a1 #i1 #i4 ,so here we have a similar negative check1 for a number cannot be triangular 7that is1 7n n7_/#b7 if its digital root be #b1 #d1 #e1 #g1 or #h4 #gf4--,,the ,,barrel ,,of ,,beer4 ,a man bought an odd lot of wine in barrels and one barrel containing beer4 ,these are shown in the illustration1 marked with the number of gallons that each barrel contained4 ,he sold a quantity of the wine to one man and twice the quantity to another1 but kept the beer to himself4 ,the puzzle is to point out which barrel contains beer4 ,can you say which one it is8 ,of course1 the man sold the barrels just as he bought them1 without manipulating in any way the contents4 ,illustration3 7 #ae ,gals 7 7#ca ,gals7 7#ai ,gals7 7#bj ,gals7 7#af ,gals7 7#ah ,gals7 #gg4--;,,digits ,,and ,,squares4 ,illustration3 --------- #a #i #b --------- #c #h #d --------- #e #g #f --------- ,it will be seen in the diagram that we have so arranged the nine digits in a square that the number in the second row is twice that in the first row1 and the number in the bottom row three times that in the top row4 ,there are three other ways of arranging the digits so as to produce the same result4 ,can you find them8 #gh4--,,odd ,,and ,,even ,,digits4 ,the odd digits1 #a1 #c1 #e1 #g1 and #i1 add up #be1 while the even figures1 #b1 #d1 #f1 and #h1 only add up #bj4 ,arrange these figures so that the odd ones and the even ones add up alike4 ,complex and improper fractions and recurring decimals are not allowed4 #gi4--,,the ,,lockers ,,puzzle4 ,illustration3 ,a ,b ,c -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- ------------------ ------------------ ------------------ ,a man had in his office three cupboards1 each containing nine lockers1 as shown in the diagram4 ,he told his clerk to place a different one-figure number on each locker of cupboard ,a1 and to do the same in the case of ,b1 and of ,c4 ,as we are here allowed to call nought a digit1 and he was not prohibited from using nought as a number1 he clearly had the option of omitting any one of ten digits from each cupboard4 ,now1 the employer did not say the lockers were to be numbered in any numerical order1 and he was surprised to find1 when the work was done1 that the figures had apparently been mixed up indiscriminately4 ,calling upon his clerk for an explanation1 the eccentric lad stated that the notion had occurred to him so to arrange the figures that in each case they formed a simple addition sum1 the two upper rows of figures producing the sum in the lowest row4 ,but the most surprising point was this3 that he had so arranged them that the addition in ,a gave the smallest possible sum1 that the addition in ,c gave the largest possible sum1 and that all the nine digits in the three totals were different4 ,the puzzle is to show how this could be done4 ,no decimals are allowed and the nought may not appear in the hundreds place4 #hj4--,,the ,,three ,,groups4 ,there appeared in ,nouvelles ,annales de ,mathmatiques the following puzzle as a modification of one of my ,canterbury ,puzzles4 ,arrange the nine digits in three groups of two1 three1 and four digits1 so that the first two numbers when multiplied together make the third4 ,thus1 #ab #dhc #e1gif4 ,i now also propose to include the cases where there are one1 four1 and four digits1 such as #d #a1gch #f1ieb4 ,can you find all the possible solutions in both cases8 #ha4--,,the ,,nine ,,counters4 ,illustration3 7#a77e77h7 7#g77i7 7#b77c7 7#d77f7 ,i have nine counters1 each bearing one of the nine digits1 #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h and #i4 ,i arranged them on the table in two groups1 as shown in the illustration1 so as to form two multiplication sums1 and found that both sums gave the same product4 ,you will find that #aeh multiplied by #bc is #c1fcd1 and that #gi multiplied by #df is also #c1fcd4 ,now1 the puzzle ,i propose is to rearrange the counters so as to get as large a product as possible4 ,what is the best way of placing them8 ,remember both groups must multiply to the same amount1 and there must be three counters multiplied by two in one case1 and two multiplied by two counters in the other1 just as at present4 #hb4--,,the ,,ten ,,counters4 ,in this case we use the nought in addition to the #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i4 ,the puzzle is1 as in the last case1 so to arrange the ten counters that the products of the two multiplications shall be the same1 and you may here have one or more figures in the multiplier1 as you choose4 ,the above is a very easy feat2 but it is also required to find the two arrangements giving pairs of the highest and lowest products possible4 ,of course every counter must be used1 and the cipher may not be placed to the left of a row of figures where it would have no effect4 ,vulgar fractions or decimals are not allowed4 #hc4--;,,digital ,,multiplication4 ,here is another entertaining problem with the nine digits1 the nought being excluded4 ,using each figure once1 and only once1 we can form two multiplication sums that have the same product1 and this may be done in many ways4 ,for example1 #g #feh and #ad #cbi contain all the digits once1 and the product in each case is the same--#d1fjf4 ,now1 it will be seen that the sum of the digits in the product is #af1 which is neither the highest nor the lowest sum so obtainable4 ,can you find the solution of the problem that gives the lowest possible sum of digits in the common product8 ,also that which gives the highest possible sum8 #hd4--,,the ,,pierrot',s ,,puzzle4 ,illustration ,the ,pierrot in the illustration is standing in a posture that represents the sign of multiplication4 ,he is indicating the peculiar fact that #ae multiplied by #ic produces exactly the same figures 7#a1cie71 differently arranged4 ,the puzzle is to take any four digits you like 7all different7 and similarly arrange them so that the number formed on one side of the ,pierrot when multiplied by the number on the other side shall produce the same figures4 ,there are very few ways of doing it1 and ,i shall give all the cases possible4 ,can you find them all8 ,you are allowed to put two figures on each side of the ,pierrot as in the example shown1 or to place a single figure on one side and three figures on the other4 ,if we only used three digits instead of four1 the only possible ways are these3 #c multiplied by #ea equals #aec1 and #f multiplied by #ba equals #abf4 #he4--,,the ,,cab ,,numbers4 ,a ,london policeman one night saw two cabs drive off in opposite directions under suspicious circumstances4 ,this officer was a particularly careful and wide-awake man1 and he took out his pocket-book to make an entry of the numbers of the cabs1 but discovered that he had lost his pencil4 ,luckily1 however1 he found a small piece of chalk1 with which he marked the two numbers on the gateway of a wharf close by4 ,when he returned to the same spot on his beat he stood and looked again at the numbers1 and noticed this peculiarity1 that all the nine digits 7no nought7 were used and that no figure was repeated1 but that if he multiplied the two numbers together they again produced the nine digits1 all once1 and once only4 ,when one of the clerks arrived at the wharf in the early morning1 he observed the chalk marks and carefully rubbed them out4 ,as the policeman could not remember them1 certain mathematicians were then consulted as to whether there was any known method for discovering all the pairs of numbers that have the peculiarity that the officer had noticed2 but they knew of none4 ,the investigation1 however1 was interesting1 and the following question out of many was proposed3 ,what two numbers1 containing together all the nine digits1 will1 when multiplied together1 produce another number 7the highest possible7 containing also all the nine digits8 ,the nought is not allowed anywhere4 #hf4--,,queer ,,multiplication4 ,if ,i multiply #ea1bdi1hgf by #c 7thus using all the nine digits once1 and once only71 ,i get #aec1gdi1fbh 7which again contains all the nine digits once74 ,similarly1 if ,i multiply #af1ehc1gdb by #i the result is #adi1bec1fgh1 where in each case all the nine digits are used4 ,now1 take #f as your multiplier and try to arrange the remaining eight digits so as to produce by multiplication a number containing all nine once1 and once only4 ,you will find it far from easy1 but it can be done4 #hg4--,,the ,,number-,,checks ,,puzzle4 ,illustration ,where a large number of workmen are employed on a building it is customary to provide every man with a little disc bearing his number4 ,these are hung on a board by the men as they arrive1 and serve as a check on punctuality4 ,now1 ,i once noticed a foreman remove a number of these checks from his board and place them on a split-ring which he carried in his pocket4 ,this at once gave me the idea for a good puzzle4 ,in fact1 ,i will confide to my readers that this is just how ideas for puzzles arise4 ,you cannot really create an idea3 it happens--and you have to be on the alert to seize it when it does so happen4 ,it will be seen from the illustration that there are ten of these checks on a ring1 numbered #a to #i and #j4 ,the puzzle is to divide them into three groups without taking any off the ring1 so that the first group multiplied by the second makes the third group4 ,for example1 we can divide them into the three groups1 #b--h #i #j #g--a #e #d #f #c1 by bringing the #f and the #c round to the #d1 but unfortunately the first two when multiplied together do not make the third4 ,can you separate them correctly8 ,of course you may have as many of the checks as you like in any group4 ,the puzzle calls for some ingenuity1 unless you have the luck to hit on the answer by chance4 #hh4--;,,digital ,,division4 ,it is another good puzzle so to arrange the nine digits 7the nought excluded7 into two groups so that one group when divided by the other produces a given number without remainder4 ,for example1 #a #c #d #e #h divided by #f #g #b #i gives #b4 ,can the reader find similar arrangements producing #c1 #d1 #e1 #f1 #g1 #h1 and #i respectively8 ,also1 can he find the pairs of smallest possible numbers in each case8 ,thus1 #a #d #f #e #h divided by #g #c #b #i is just as correct for #b as the other example we have given1 but the numbers are higher4 #hi4--;,,adding ,,the ,,digits4 ,if ,i write the sum of money1 #ihg1 #es4 #d;d41 and add up the digits1 they sum to #cf4 ,no digit has thus been used a second time in the amount or addition4 ,this is the largest amount possible under the conditions4 ,now find the smallest possible amount1 pounds1 shillings1 pence1 and farthings being all represented4 ,you need not use more of the nine digits than you choose1 but no digit may be repeated throughout4 ,the nought is not allowed4 #ij4--,,the ,,century ,,puzzle4 ,can you write #ajj in the form of a mixed number1 using all the nine digits once1 and only once8 ,the late distinguished ,french mathematician1 ,edouard ,lucas1 found seven different ways of doing it1 and expressed his doubts as to there being any other ways4 ,as a matter of fact there are just eleven ways and no more4 ,here is one of them1 #iaegdb_/fch4 ,nine of the other ways have similarly two figures in the integral part of the number1 but the eleventh expression has only one figure there4 ,can the reader find this last form8 #ia4--,,more ,,mixed ,,fractions4 ,when ,i first published my solution to the last puzzle1 ,i was led to attempt the expression of all numbers in turn up to #ajj by a mixed fraction containing all the nine digits4 ,here are twelve numbers for the reader to try his hand at3 #ac1 #ad1 #ae1 #af1 #ah1 #bj1 #bg1 #cf1 #dj1 #fi1 #gb1 #id4 ,use every one of the nine digits once1 and only once1 in every case4 #ib4--;,,digital ,,square ,,numbers4 ,here are the nine digits so arranged that they form four square numbers3 #i1 #ha1 #cbd1 #egf4 ,now1 can you put them all together so as to form a single square number--7,i7 the smallest possible1 and 7,,ii7 the largest possible8 #ic4--,,the ,,mystic ,,eleven4 ,can you find the largest possible number containing any nine of the ten digits 7calling nought a digit7 that can be divided by #aa without a remainder8 ,can you also find the smallest possible number produced in the same way that is divisible by #aa8 ,here is an example1 where the digit #e has been omitted3 #hifgdcjab4 ,this number contains nine of the digits and is divisible by #aa1 but it is neither the largest nor the smallest number that will work4 #id4--,,the ,,digital ,,century4 #a #b #c #d #e #f #g #h #i #ajj4 ,it is required to place arithmetical signs between the nine figures so that they shall equal #ajj4 ,of course1 you must not alter the present numerical arrangement of the figures4 ,can you give a correct solution that employs 7#a7 the fewest possible signs1 and 7#b7 the fewest possible separate strokes or dots of the pen8 ,that is1 it is necessary to use as few signs as possible1 and those signs should be of the simplest form4 ,the signs of addition and multiplication 7 and 7 will thus count as two strokes1 the sign of subtraction 7-7 as one stroke1 the sign of division 77 as three1 and so on4 #ie4--,,the ,,four ,,sevens4 ,illustration ,in the illustration ,professor ,rackbrane is seen demonstrating one of the little posers with which he is accustomed to entertain his class4 ,he believes that by taking his pupils off the beaten tracks he is the better able to secure their attention1 and to induce original and ingenious methods of thought4 ,he has1 it will be seen1 just shown how four #e's may be written with simple arithmetical signs so as to represent #ajj4 ,every juvenile reader will see at a glance that his example is quite correct4 ,now1 what he wants you to do is this3 ,arrange four #g's 7neither more nor less7 with arithmetical signs so that they shall represent #ajj4 ,if he had said we were to use four #i's we might at once have written #iii_/i1 but the four #g's call for rather more ingenuity4 ,can you discover the little trick8 #if4--,,the ,,dice ,,numbers4 ,illustration ,i have a set of four dice1 not marked with spots in the ordinary way1 but with ,arabic figures1 as shown in the illustration4 ,each die1 of course1 bears the numbers #a to #f4 ,when put together they will form a good many1 different numbers4 ,as represented they make the number #abdf4 ,now1 if ,i make all the different four-figure numbers that are possible with these dice 7never putting the same figure more than once in any number71 what will they all add up to8 ,you are allowed to turn the #f upside down1 so as to represent a #i4 ,i do not ask1 or expect1 the reader to go to all the labour of writing out the full list of numbers and then adding them up4 ,life is not long enough for such wasted energy4 ,can you get at the answer in any other way8 ,,various ,,arithmetical ,,and ,,algebraical ,,problems4 ,variety's the very spice of life1 ,that gives it all its flavour4 ,,cowper3 ,the ,task4 #ig4--,,the ,,spot ,,on ,,the ,,table4 ,a boy1 recently home from school1 wished to give his father an exhibition of his precocity4 ,he pushed a large circular table into the corner of the room1 as shown in the illustration1 so that it touched both walls1 and he then pointed to a spot of ink on the extreme edge4 ,illustration ,here is a little puzzle for you1 pater1 said the youth4 ,that spot is exactly eight inches from one wall and nine inches from the other4 ,can you tell me the diameter of the table without measuring it8 ,the boy was overheard to tell a friend1 ,it fairly beat the guv'nor2 but his father is known to have remarked to a ,city acquaintance that he solved the thing in his head in a minute4 ,i often wonder which spoke the truth4 #ih4--;,,academic ,,courtesies4 ,in a certain mixed school1 where a special feature was made of the inculcation of good manners1 they had a curious rule on assembling every morning4 ,there were twice as many girls as boys4 ,every girl made a bow to every other girl1 to every boy1 and to the teacher4 ,every boy made a bow to every other boy1 to every girl1 and to the teacher4 ,in all there were nine hundred bows made in that model academy every morning4 ,now1 can you say exactly how many boys there were in the school8 ,if you are not very careful1 you are likely to get a good deal out in your calculation4 #ii4--,,the ,,thirty-,,three ,,pearls4 ,illustration ,a man ,i know1 said ,teddy ,nicholson at a certain family party1 possesses a string of thirty-three pearls4 ,the middle pearl is the largest and best of all1 and the others are so selected and arranged that1 starting from one end1 each successive pearl is worth #ajj more than the preceding one1 right up to the big pearl4 ,from the other end the pearls increase in value by #aej up to the large pearl4 ,the whole string is worth #fe1jjj4 ,what is the value of that large pearl8 ,pearls and other articles of clothing1 said ,uncle ,walter1 when the price of the precious gem had been discovered1 remind me of ,adam and ,eve4 ,authorities1 you may not know1 differ as to the number of apples that were eaten by ,adam and ,eve4 ,it is the opinion of some that ,eve #h 7ate7 and ,adam #b 7too71 a total of #aj only4 ,but certain mathematicians have figured it out differently1 and hold that ,eve #h and ,adam a total of #af4 ,yet the most recent investigators think the above figures entirely wrong1 for if ,eve #h and ,adam #h #b1 the total must be #ij4 ,well1 said ,harry1 it seems to me that if there were giants in those days1 probably ,eve #h #a and ,adam #h #b1 which would give a total of #afc4 ,i am not at all satisfied1 said ,maud4 ,it seems to me that if ,eve #h #a and ,adam #h #a #b1 they together consumed #hic4 ,i am sure you are all wrong1 insisted ,mr4 ,wilson1 for ,i consider that ,eve #h #a #d ,adam1 and ,adam #h #a #b #d ,eve1 so we get a total of #h1ich4 ,but1 look here1 broke in ,herbert4 ,if ,eve #h #a #d ,adam and ,adam #h #a #b #d #b oblige ,eve1 surely the total must have been #hb1jef6 ,at this point ,uncle ,walter suggested that they might let the matter rest4 ,he declared it to be clearly what mathematicians call an indeterminate problem4 #ajj4--,,the ,,labourer',s ,,puzzle4 ,professor ,rackbrane1 during one of his rambles1 chanced to come upon a man digging a deep hole4 ,good morning1 he said4 ,how deep is that hole8 ,guess1 replied the labourer4 ,my height is exactly five feet ten inches4 ,how much deeper are you going8 said the professor4 ,i am going twice as deep1 was the answer1 and then my head will be twice as far below ground as it is now above ground4 ,rackbrane now asks if you could tell how deep that hole would be when finished4 #aja4--,,the ,,trusses ,,of ,,hay4 ,farmer ,tompkins had five trusses of hay1 which he told his man ,hodge to weigh before delivering them to a customer4 ,the stupid fellow weighed them two at a time in all possible ways1 and informed his master that the weights in pounds were #aaj1 #aab1 #aac1 #aad1 #aae1 #aaf1 #aag1 #aah1 #abj1 and #aba4 ,now1 how was ,farmer ,tompkins to find out from these figures how much every one of the five trusses weighed singly8 ,the reader may at first think that he ought to be told which pair is which pair1 or something of that sort1 but it is quite unnecessary4 ,can you give the five correct weights8 #ajb4--,,mr4 ,,gubbins ,,in ,a ,,fog4 ,mr4 ,gubbins1 a diligent man of business1 was much inconvenienced by a ,london fog4 ,the electric light happened to be out of order and he had to manage as best he could with two candles4 ,his clerk assured him that though both were of the same length one candle would burn for four hours and the other for five hours4 ,after he had been working some time he put the candles out as the fog had lifted1 and he then noticed that what remained of one candle was exactly four times the length of what was left of the other4 ,when he got home that night ,mr4 ,gubbins1 who liked a good puzzle1 said to himself1 ,of course it is possible to work out just how long those two candles were burning to-day4 ,i'll have a shot at it4 ,but he soon found himself in a worse fog than the atmospheric one4 ,could you have assisted him in his dilemma8 ,how long were the candles burning8 #ajc4--,,painting ,,the ,,lamp-,,posts4 ,tim ,murphy and ,pat ,donovan were engaged by the local authorities to paint the lamp-posts in a certain street4 ,tim1 who was an early riser1 arrived first on the job1 and had painted three on the south side when ,pat turned up and pointed out that ,tim's contract was for the north side4 ,so ,tim started afresh on the north side and ,pat continued on the south4 ,when ,pat had finished his side he went across the street and painted six posts for ,tim1 and then the job was finished4 ,as there was an equal number of lamp-posts on each side of the street1 the simple question is3 ,which man painted the more lamp-posts1 and just how many more8 #ajd4--;,,catching ,,the ,,thief4 ,now1 constable1 said the defendant's counsel in cross-examination1 you say that the prisoner was exactly twenty-seven steps ahead of you when you started to run after him8 ,yes1 sir4 ,and you swear that he takes eight steps to your five8 ,that is so4 ,then ,i ask you1 constable1 as an intelligent man1 to explain how you ever caught him1 if that is the case8 ,well1 you see1 ,i have got a longer stride4 ,in fact1 two of my steps are equal in length to five of the prisoner's4 ,if you work it out1 you will find that the number of steps ,i required would bring me exactly to the spot where ,i captured him4 ,here the foreman of the jury asked for a few minutes to figure out the number of steps the constable must have taken4 ,can you also say how many steps the officer needed to catch the thief8 #aje4--,,the ,,parish ,,council ,,election4 ,here is an easy problem for the novice4 ,at the last election of the parish council of ,tittlebury-in-the-,marsh there were twenty-three candidates for nine seats4 ,each voter was qualified to vote for nine of these candidates or for any less number4 ,one of the electors wants to know in just how many different ways it was possible for him to vote4 #ajf4--,,the ,,muddletown ,,election4 ,at the last ,parliamentary election at ,muddletown #e1dgc votes were polled4 ,the ,liberal was elected by a majority of #ah over the ,conservative1 by #adf over the ,independent1 and by #ege over the ,socialist4 ,can you give a simple rule for figuring out how many votes were polled for each candidate8 #ajg4--,,the ,,suffragists' ,,meeting4 ,at a recent secret meeting of ,suffragists a serious difference of opinion arose4 ,this led to a split1 and a certain number left the meeting4 ,i had half a mind to go myself1 said the chair-woman1 and if ,i had done so1 two-thirds of us would have retired4 ,true1 said another member2 but if ,i had persuaded my friends ,mrs4 ,wild and ,christine ,armstrong to remain we should only have lost half our number4 ,can you tell how many were present at the meeting at the start8 #ajh4--,,the ,,leap-,,year ,,ladies4 ,last leap-year ladies lost no time in exercising the privilege of making proposals of marriage4 ,if the figures that reached me from an occult source are correct1 the following represents the state of affairs in this country4 ,a number of women proposed once each1 of whom one-eighth were widows4 ,in consequence1 a number of men were to be married of whom one-eleventh were widowers4 ,of the proposals made to widowers1 one-fifth were declined4 ,all the widows were accepted4 ,thirty-five forty-fourths of the widows married bachelors4 ,one thousand two hundred and twenty-one spinsters were declined by bachelors4 ,the number of spinsters accepted by bachelors was seven times the number of widows accepted by bachelors4 ,those are all the particulars that ,i was able to obtain4 ,now1 how many women proposed8 #aji4--,,the ,,great ,,scramble4 ,after dinner1 the five boys of a household happened to find a parcel of sugar-plums4 ,it was quite unexpected loot1 and an exciting scramble ensued1 the full details of which ,i will recount with accuracy1 as it forms an interesting puzzle4 ,you see1 ,andrew managed to get possession of just two-thirds of the parcel of sugar-plums4 ,bob at once grabbed three-eighths of these1 and ,charlie managed to seize three-tenths also4 ,then young ,david dashed upon the scene1 and captured all that ,andrew had left1 except one-seventh1 which ,edgar artfully secured for himself by a cunning trick4 ,now the fun began in real earnest1 for ,andrew and ,charlie jointly set upon ,bob1 who stumbled against the fender and dropped half of all that he had1 which were equally picked up by ,david and ,edgar1 who had crawled under a table and were waiting4 ,next1 ,bob sprang on ,charlie from a chair1 and upset all the latter's collection on to the floor4 ,of this prize ,andrew got just a quarter1 ,bob gathered up one-third1 ,david got two-sevenths1 while ,charlie and ,edgar divided equally what was left of that stock4 ,illustration ,they were just thinking the fray was over when ,david suddenly struck out in two directions at once1 upsetting three-quarters of what ,bob and ,andrew had last acquired4 ,the two latter1 with the greatest difficulty1 recovered five-eighths of it in equal shares1 but the three others each carried off one-fifth of the same4 ,every sugar-plum was now accounted for1 and they called a truce1 and divided equally amongst them the remainder of the parcel4 ,what is the smallest number of sugar-plums there could have been at the start1 and what proportion did each boy obtain8 #aaj4--,,the ,,abbot',s ,,puzzle4 ,the first ,english puzzlist whose name has come down to us was a ,yorkshireman--no other than ,alcuin1 ,abbot of ,canterbury 7,a4,d4 #gce-hjd74 ,here is a little puzzle from his works1 which is at least interesting on account of its antiquity4 ,if #ajj bushels of corn were distributed among #ajj people in such a manner that each man received three bushels1 each woman two1 and each child half a bushel1 how many men1 women1 and children were there8 ,now1 there are six different correct answers1 if we exclude a case where there would be no women4 ,but let us say that there were just five times as many women as men1 then what is the correct solution8 #aaa4--,,reaping ,,the ,,corn4 ,a farmer had a square cornfield4 ,the corn was all ripe for reaping1 and1 as he was short of men1 it was arranged that he and his son should share the work between them4 ,the farmer first cut one rod wide all round the square1 thus leaving a smaller square of standing corn in the middle of the field4 ,now1 he said to his son1 ,i have cut my half of the field1 and you can do your share4 ,the son was not quite satisfied as to the proposed division of labour1 and as the village schoolmaster happened to be passing1 he appealed to that person to decide the matter4 ,he found the farmer was quite correct1 provided there was no dispute as to the size of the field1 and on this point they were agreed4 ,can you tell the area of the field1 as that ingenious schoolmaster succeeded in doing8 #aab4--;,a ,,puzzling ,,legacy4 ,a man left a hundred acres of land to be divided among his three sons--,alfred1 ,benjamin1 and ,charles--in the proportion of one-third1 one-fourth1 and one-fifth respectively4 ,but ,charles died4 ,how was the land to be divided fairly between ,alfred and ,benjamin8 #aac4--,,the ,,torn ,,number4 ,illustration ,i had the other day in my possession a label bearing the number #c #j #b #e in large figures4 ,this got accidentally torn in half1 so that #c #j was on one piece and #b #e on the other1 as shown on the illustration4 ,on looking at these pieces ,i began to make a calculation1 scarcely conscious of what ,i was doing1 when ,i discovered this little peculiarity4 ,if we add the #c #j and the #b #e together and square the sum we get as the result the complete original number on the label6 ,thus1 #cj added to #be is #ee1 and #ee multiplied by #ee is #cjbe4 ,curious1 is it not8 ,now1 the puzzle is to find another number1 composed of four figures1 all different1 which may be divided in the middle and produce the same result4 #aad4--;,,curious ,,numbers4 ,the number #dh has this peculiarity1 that if you add #a to it the result is a square number 7#di1 the square of #g71 and if you add #a to its half1 you also get a square number 7#be1 the square of #e74 ,now1 there is no limit to the numbers that have this peculiarity1 and it is an interesting puzzle to find three more of them--the smallest possible numbers4 ,what are they8 #aae4--;,a ,,printer',s ,,error4 ,in a certain article a printer had to set up the figures #edbc1 which1 of course1 means that the fourth power of #e 7#fbe7 is to be multiplied by the cube of #b 7#h71 the product of which is #e1jjj4 ,but he printed #edbc as #e #d #b #c1 which is not correct4 ,can you place four digits in the manner shown1 so that it will be equally correct if the printer sets it up aright or makes the same blunder8 #aaf4--,,the ,,converted ,,miser4 ,mr4 ,jasper ,bullyon was one of the very few misers who have ever been converted to a sense of their duty towards their less fortunate fellow-men4 ,one eventful night he counted out his accumulated wealth1 and resolved to distribute it amongst the deserving poor4 ,he found that if he gave away the same number of pounds every day in the year1 he could exactly spread it over a twelvemonth without there being anything left over2 but if he rested on the ,sundays1 and only gave away a fixed number of pounds every weekday1 there would be one sovereign left over on ,new ,year's ,eve4 ,now1 putting it at the lowest possible1 what was the exact number of pounds that he had to distribute8 ,could any question be simpler8 ,a sum of pounds divided by one number of days leaves no remainder1 but divided by another number of days leaves a sovereign over4 ,that is all2 and yet1 when you come to tackle this little question1 you will be surprised that it can become so puzzling4 #aag4--;,a ,,fence ,,problem4 ,illustration ,the practical usefulness of puzzles is a point that we are liable to overlook4 ,yet1 as a matter of fact1 ,i have from time to time received quite a large number of letters from individuals who have found that the mastering of some little principle upon which a puzzle was built has proved of considerable value to them in a most unexpected way4 ,indeed1 it may be accepted as a good maxim that a puzzle is of little real value unless1 as well as being amusing and perplexing1 it conceals some instructive and possibly useful feature4 ,it is1 however1 very curious how these little bits of acquired knowledge dovetail into the occasional requirements of everyday life1 and equally curious to what strange and mysterious uses some of our readers seem to apply them4 ,what1 for example1 can be the object of ,mr4 ,wm4 ,oxley1 who writes to me all the way from ,iowa1 in wishing to ascertain the dimensions of a field that he proposes to enclose1 containing just as many acres as there shall be rails in the fence8 ,the man wishes to fence in a perfectly square field which is to contain just as many acres as there are rails in the required fence4 ,each hurdle1 or portion of fence1 is seven rails high1 and two lengths would extend one pole 7#af ft473 that is to say1 there are fourteen rails to the pole1 lineal measure4 ,now1 what must be the size of the field8 #aah4--;,,circling ,,the ,,squares4 ,illustration ,the puzzle is to place a different number in each of the ten squares so that the sum of the squares of any two adjacent numbers shall be equal to the sum of the squares of the two numbers diametrically opposite to them4 ,the four numbers placed1 as examples1 must stand as they are4 ,the square of #af is #bef1 and the square of #b is #d4 ,add these together1 and the result is #bfj4 ,also--the square of #ad is #aif1 and the square of #h is #fd4 ,these together also make #bfj4 ,now1 in precisely the same way1 ,b and ,c should be equal to ,g and ,h 7the sum will not necessarily be #bfj71 ,a and ,k to ,f and ,e1 ,h and ,i to ,c and ,d1 and so on1 with any two adjoining squares in the circle4 ,all you have to do is to fill in the remaining six numbers4 ,fractions are not allowed1 and ,i shall show that no number need contain more than two figures4 #aai4--,,rackbrane',s ,,little ,,loss4 ,professor ,rackbrane was spending an evening with his old friends1 ,mr4 and ,mrs4 ,potts1 and they engaged in some game 7he does not say what game7 of cards4 ,the professor lost the first game1 which resulted in doubling the money that both ,mr4 and ,mrs4 ,potts had laid on the table4 ,the second game was lost by ,mrs4 ,potts1 which doubled the money then held by her husband and the professor4 ,curiously enough1 the third game was lost by ,mr4 ,potts1 and had the effect of doubling the money then held by his wife and the professor4 ,it was then found that each person had exactly the same money1 but the professor had lost five shillings in the course of play4 ,now1 the professor asks1 what was the sum of money with which he sat down at the table8 ,can you tell him8 #abj4--,,the ,,farmer ,,and ,,his ,,sheep4 ,illustration ,farmer ,longmore had a curious aptitude for arithmetic1 and was known in his district as the mathematical farmer4 ,the new vicar was not aware of this fact when1 meeting his worthy parishioner one day in the lane1 he asked him in the course of a short conversation1 ,now1 how many sheep have you altogether8 ,he was therefore rather surprised at ,longmore's answer1 which was as follows3 ,you can divide my sheep into two different parts1 so that the difference between the two numbers is the same as the difference between their squares4 ,maybe1 ,mr4 ,parson1 you will like to work out the little sum for yourself4 ,can the reader say just how many sheep the farmer had8 ,supposing he had possessed only twenty sheep1 and he divided them into the two parts #ab and #h4 ,now1 the difference between their squares1 #add and #fd1 is #hj4 ,so that will not do1 for #d and #hj are certainly not the same4 ,if you can find numbers that work out correctly1 you will know exactly how many sheep ,farmer ,longmore owned4 #aba4--;,,heads ,,or ,,tails4 ,crooks1 an inveterate gambler1 at ,goodwood recently said to a friend1 ,i'll bet you half the money in my pocket on the toss of a coin--heads ,i win1 tails ,i lose4 ,the coin was tossed and the money handed over4 ,he repeated the offer again and again1 each time betting half the money then in his possession4 ,we are not told how long the game went on1 or how many times the coin was tossed1 but this we know1 that the number of times that ,crooks lost was exactly equal to the number of times that he won4 ,now1 did he gain or lose by this little venture8 #abb4--,,the ,,see-,,saw ,,puzzle4 ,necessity is1 indeed1 the mother of invention4 ,i was amused the other day in watching a boy who wanted to play see-saw and1 in his failure to find another child to share the sport with him1 had been driven back upon the ingenious resort of tying a number of bricks to one end of the plank to balance his weight at the other4 ,as a matter of fact1 he just balanced against sixteen bricks1 when these were fixed to the short end of plank1 but if he fixed them to the long end of plank he only needed eleven as balance4 ,now1 what was that boy's weight1 if a brick weighs equal to a three-quarter brick and three-quarters of a pound8 #abc4--;,a ,,legal ,,difficulty4 ,a client of mine1 said a lawyer1 was on the point of death when his wife was about to present him with a child4 ,i drew up his will1 in which he settled two-thirds of his estate upon his son 7if it should happen to be a boy7 and one-third on the mother4 ,but if the child should be a girl1 then two-thirds of the estate should go to the mother and one-third to the daughter4 ,as a matter of fact1 after his death twins were born--a boy and a girl4 ,a very nice point then arose4 ,how was the estate to be equitably divided among the three in the closest possible accordance with the spirit of the dead man's will8 #abd4--;,a ,,question ,,of ,,definition4 ,my property is exactly a mile square1 said one landowner to another4 ,curiously enough1 mine is a square mile1 was the reply4 ,then there is no difference8 ,is this last statement correct8 #abe4--,,the ,,miners' ,,holiday4 ,seven coal-miners took a holiday at the seaside during a big strike4 ,six of the party spent exactly half a sovereign each1 but ,bill ,harris was more extravagant4 ,bill spent three shillings more than the average of the party4 ,what was the actual amount of ,bill's expenditure8 #abf4--,,simple ,,multiplication4 ,if we number six cards #a1 #b1 #d1 #e1 #g1 and #h1 and arrange them on the table in this order3-- #a #d #b #h #e #g ,we can demonstrate that in order to multiply by #c all that is necessary is to remove the #a to the other end of the row1 and the thing is done4 ,the answer is #dbhega4 ,can you find a number that1 when multiplied by #c and divided by #b1 the answer will be the same as if we removed the first card 7which in this case is to be a #c7 ,from the beginning of the row to the end8 #abg4--,,simple ,,division4 ,sometimes a very simple question in elementary arithmetic will cause a good deal of perplexity4 ,for example1 ,i want to divide the four numbers1 #gja1 #a1jei1 #a1dag1 and #b1cab1 by the largest number possible that will leave the same remainder in every case4 ,how am ,i to set to work ,of course1 by a laborious system of trial one can in time discover the answer1 but there is quite a simple method of doing it if you can only find it4 #abh4--;,a ,,problem ,,in ,,squares4 ,we possess three square boards4 ,the surface of the first contains five square feet more than the second1 and the second contains five square feet more than the third4 ,can you give exact measurements for the sides of the boards8 ,if you can solve this little puzzle1 then try to find three squares in arithmetical progression1 with a common difference of #g and also of #ac4 #abi4--,,the ,,battle ,,of ,,hastings4 ,all historians know that there is a great deal of mystery and uncertainty concerning the details of the ever-memorable battle on that fatal day1 ,october #ad1 #ajff4 ,my puzzle deals with a curious passage in an ancient monkish chronicle that may never receive the attention that it deserves1 and if ,i am unable to vouch for the authenticity of the document it will none the less serve to furnish us with a problem that can hardly fail to interest those of my readers who have arithmetical predilections4 ,here is the passage in question4 ,the men of ,harold stood well together1 as their wont was1 and formed sixty and one squares1 with a like number of men in every square thereof1 and woe to the hardy ,norman who ventured to enter their redoubts2 for a single blow of a ,saxon war-hatchet would break his lance and cut through his coat of mail4444 ,when ,harold threw himself into the fray the ,saxons were one mighty square of men1 shouting the battle-cries1 ',ut6' ',olicrosse6' ',godemit6' ,now1 ,i find that all the contemporary authorities agree that the ,saxons did actually fight in this solid order4 ,for example1 in the ,carmen de ,bello ,hastingensi1 a poem attributed to ,guy1 ,bishop of ,amiens1 living at the time of the battle1 we are told that the ,saxons stood fixed in a dense mass1 and ,henry of ,huntingdon records that they were like unto a castle1 impenetrable to the ,normans2 while ,robert ,wace1 a century after1 tells us the same thing4 ,so in this respect my newly-discovered chronicle may not be greatly in error4 ,but ,i have reason to believe that there is something wrong with the actual figures4 ,let the reader see what he can make of them4 ,the number of men would be sixty-one times a square number2 but when ,harold himself joined in the fray they were then able to form one large square4 ,what is the smallest possible number of men there could have been8 ,in order to make clear to the reader the simplicity of the question1 ,i will give the lowest solutions in the case of #fj and #fb1 the numbers immediately preceding and following #fa4 ,they are #fj #d #a #ca1 and #fb #h #a #fc4 ,that is1 #fj squares of #af men each would be #ifj men1 and when ,harold joined them they would be #ifa in number1 and so form a square with #ca men on every side4 ,similarly in the case of the figures ,i have given for #fb4 ,now1 find the lowest answer for #fa4 #acj4--,,the ,,sculptor',s ,,problem4 ,an ancient sculptor was commissioned to supply two statues1 each on a cubical pedestal4 ,it is with these pedestals that we are concerned4 ,they were of unequal sizes1 as will be seen in the illustration1 and when the time arrived for payment a dispute arose as to whether the agreement was based on lineal or cubical measurement4 ,but as soon as they came to measure the two pedestals the matter was at once settled1 because1 curiously enough1 the number of lineal feet was exactly the same as the number of cubical feet4 ,the puzzle is to find the dimensions for two pedestals having this peculiarity1 in the smallest possible figures4 ,you see1 if the two pedestals1 for example1 measure respectively #c ft4 and #a ft4 on every side1 then the lineal measurement would be #d ft4 and the cubical contents #bh ft41 which are not the same1 so these measurements will not do4 ,illustration #aca4--,,the ,,spanish ,,miser4 ,there once lived in a small town in ,new ,castile a noted miser named ,don ,manuel ,rodriguez4 ,his love of money was only equalled by a strong passion for arithmetical problems4 ,these puzzles usually dealt in some way or other with his accumulated treasure1 and were propounded by him solely in order that he might have the pleasure of solving them himself4 ,unfortunately very few of them have survived1 and when travelling through ,spain1 collecting material for a proposed work on ,the ,spanish ,onion as a ,cause of ,national ,decadence1 ,i only discovered a very few4 ,one of these concerns the three boxes that appear in the accompanying authentic portrait4 ,illustration ,each box contained a different number of golden doubloons4 ,the difference between the number of doubloons in the upper box and the number in the middle box was the same as the difference between the number in the middle box and the number in the bottom box4 ,and if the contents of any two of the boxes were united they would form a square number4 ,what is the smallest number of doubloons that there could have been in any one of the boxes8 #acb4--,,the ,,nine ,,treasure ,,boxes4 ,the following puzzle will illustrate the importance on occasions of being able to fix the minimum and maximum limits of a required number4 ,this can very frequently be done4 ,for example1 it has not yet been ascertained in how many different ways the knight's tour can be performed on the chess board2 but we know that it is fewer than the number of combinations of #afh things taken #fc at a time and is greater than #ca1jed1add--;for the latter is the number of routes of a particular type4 ,or1 to take a more familiar case1 if you ask a man how many coins he has in his pocket1 he may tell you that he has not the slightest idea4 ,but on further questioning you will get out of him some such statement as the following3 ,yes1 ,i am positive that ,i have more than three coins1 and equally certain that there are not so many as twenty-five4 ,now1 the knowledge that a certain number lies between #b and #ab in my puzzle will enable the solver to find the exact answer2 without that information there would be an infinite number of answers1 from which it would be impossible to select the correct one4 ,this is another puzzle received from my friend ,don ,manuel ,rodriguez1 the cranky miser of ,new ,castile4 ,on ,new ,year's ,eve in #ahgi he showed me nine treasure boxes1 and after informing me that every box contained a square number of golden doubloons1 and that the difference between the contents of ,a and ,b was the same as between ,b and ,c1 ,d and ,e1 ,e and ,f1 ,g and ,h1 or ,h and ,i1 he requested me to tell him the number of coins in every one of the boxes4 ,at first ,i thought this was impossible1 as there would be an infinite number of different answers1 but on consideration ,i found that this was not the case4 ,i discovered that while every box contained coins1 the contents of ,a1 ,b1 ,c increased in weight in alphabetical order2 so did ,d1 ,e1 ,f2 and so did ,g1 ,h1 ,i2 but ,d or ,e need not be heavier than ,c1 nor ,g or ,h heavier than ,f4 ,it was also perfectly certain that box ,a could not contain more than a dozen coins at the outside2 there might not be half that number1 but ,i was positive that there were not more than twelve4 ,with this knowledge ,i was able to arrive at the correct answer4 ,in short1 we have to discover nine square numbers such that ,a1 ,b1 ,c2 and ,d1 ,e1 ,f2 and ,g1 ,h1 ,i are three groups in arithmetical progression1 the common difference being the same in each group1 and ,a being less than #ab4 ,how many doubloons were there in every one of the nine boxes8 #acc4--,,the ,,five ,,brigands4 ,the five ,spanish brigands1 ,alfonso1 ,benito1 ,carlos1 ,diego1 and ,esteban1 were counting their spoils after a raid1 when it was found that they had captured altogether exactly #bjj doubloons4 ,one of the band pointed out that if ,alfonso had twelve times as much1 ,benito three times as much1 ,carlos the same amount1 ,diego half as much1 and ,esteban one-third as much1 they would still have altogether just #bjj doubloons4 ,how many doubloons had each8 ,there are a good many equally correct answers to this question4 ,here is one of them3 ,a #f #ab #gb ,b #ab #c #cf ,c #ag #a #ag ,d #abj #fj ,e #de #a_/c #ae #bjj #bjj ,the puzzle is to discover exactly how many different answers there are1 it being understood that every man had something and that there is to be no fractional money--only doubloons in every case4 ,this problem1 worded somewhat differently1 was propounded by ,tartaglia 7died #aeei71 and he flattered himself that he had found one solution2 but a ,french mathematician of note 7,m4,a4 ,labosne71 in a recent work1 says that his readers will be astonished when he assures them that there are #f1fci different correct answers to the question4 ,is this so8 ,how many answers are there8 #acd4--,,the ,,banker',s ,,puzzle4 ,a banker had a sporting customer who was always anxious to wager on anything4 ,hoping to cure him of his bad habit1 he proposed as a wager that the customer would not be able to divide up the contents of a box containing only sixpences into an exact number of equal piles of sixpences4 ,the banker was first to put in one or more sixpences 7as many as he liked72 then the customer was to put in one or more 7but in his case not more than a pound in value71 neither knowing what the other put in4 ,lastly1 the customer was to transfer from the banker's counter to the box as many sixpences as the banker desired him to put in4 ,the puzzle is to find how many sixpences the banker should first put in and how many he should ask the customer to transfer1 so that he may have the best chance of winning4 #ace4--,,the ,,stonemason',s ,,problem4 ,a stonemason once had a large number of cubic blocks of stone in his yard1 all of exactly the same size4 ,he had some very fanciful little ways1 and one of his queer notions was to keep these blocks piled in cubical heaps1 no two heaps containing the same number of blocks4 ,he had discovered for himself 7a fact that is well known to mathematicians7 that if he took all the blocks contained in any number of heaps in regular order1 beginning with the single cube1 he could always arrange those on the ground so as to form a perfect square4 ,this will be clear to the reader1 because one block is a square1 #a #h #i is a square1 #a #h #bg #cf is a square1 #a #h #bg #fd #ajj is a square1 and so on4 ,in fact1 the sum of any number of consecutive cubes1 beginning always with #a1 is in every case a square number4 ,one day a gentleman entered the mason's yard and offered him a certain price if he would supply him with a consecutive number of these cubical heaps which should contain altogether a number of blocks that could be laid out to form a square1 but the buyer insisted on more than three heaps and declined to take the single block because it contained a flaw4 ,what was the smallest possible number of blocks of stone that the mason had to supply8 #acf4--,,the ,,sultan',s ,,army4 ,a certain ,sultan wished to send into battle an army that could be formed into two perfect squares in twelve different ways4 ,what is the smallest number of men of which that army could be composed8 ,to make it clear to the novice1 ,i will explain that if there were #acj men1 they could be formed into two squares in only two different ways--#ha and #di1 or #aba and #i4 ,of course1 all the men must be used on every occasion4 #acg4--;,a ,,study ,,in ,,thrift4 ,certain numbers are called triangular1 because if they are taken to represent counters or coins they may be laid out on the table so as to form triangles4 ,the number #a is always regarded as triangular1 just as #a is a square and a cube number4 ,place one counter on the table--that is1 the first triangular number4 ,now place two more counters beneath it1 and you have a triangle of three counters2 therefore #c is triangular4 ,next place a row of three more counters1 and you have a triangle of six counters2 therefore #f is triangular4 ,we see that every row of counters that we add1 containing just one more counter than the row above it1 makes a larger triangle4 ,now1 half the sum of any number and its square is always a triangular number4 ,thus half of #b #b #c2 half of #c #c #f2 half of #d #d #aj2 half of #e #e #ae2 and so on4 ,so if we want to form a triangle with #h counters on each side we shall require half of #h #h1 or #cf counters4 ,this is a pretty little property of numbers4 ,before going further1 ,i will here say that if the reader refers to the ,stonemason's ,problem 7,no4 #ace7 he will remember that the sum of any number of consecutive cubes beginning with #a is always a square1 and these form the series #a1 #c1 #f1 #aj1 etc4 ,it will now be understood when ,i say that one of the keys to the puzzle was the fact that these are always the squares of triangular numbers--that is1 the squares of #a1 #c1 #f1 #aj1 #ae1 #ba1 #bh1 etc41 any of which numbers we have seen will form a triangle4 ,every whole number is either triangular1 or the sum of two triangular numbers or the sum of three triangular numbers4 ,that is1 if we take any number we choose we can always form one1 two1 or three triangles with them4 ,the number #a will obviously1 and uniquely1 only form one triangle2 some numbers will only form two triangles 7as #b1 #d1 #aa1 etc472 some numbers will only form three triangles 7as #e1 #h1 #ad1 etc474 ,then1 again1 some numbers will form both one and two triangles 7as #f71 others both one and three triangles 7as #c and #aj71 others both two and three triangles 7as #g and #i71 while some numbers 7like #ba7 will form one1 two1 or three triangles1 as we desire4 ,now for a little puzzle in triangular numbers4 ,sandy ,mc,allister1 of ,aberdeen1 practised strict domestic economy1 and was anxious to train his good wife in his own habits of thrift4 ,he told her last ,new ,year's ,eve that when she had saved so many sovereigns that she could lay them all out on the table so as to form a perfect square1 or a perfect triangle1 or two triangles1 or three triangles1 just as he might choose to ask he would add five pounds to her treasure4 ,soon she went to her husband with a little bag of #cf in sovereigns and claimed her reward4 ,it will be found that the thirty-six coins will form a square 7with side #f71 that they will form a single triangle 7with side #h71 that they will form two triangles 7with sides #e and #f71 and that they will form three triangles 7with sides #c1 #e1 and #e74 ,in each of the four cases all the thirty-six coins are used1 as required1 and ,sandy therefore made his wife the promised present like an honest man4 ,the ,scotsman then undertook to extend his promise for five more years1 so that if next year the increased number of sovereigns that she has saved can be laid out in the same four different ways she will receive a second present2 if she succeeds in the following year she will get a third present1 and so on until she has earned six presents in all4 ,now1 how many sovereigns must she put together before she can win the sixth present8 ,what you have to do is to find five numbers1 the smallest possible1 higher than #cf1 that can be displayed in the four ways--to form a square1 to form a triangle1 to form two triangles1 and to form three triangles4 ,the highest of your five numbers will be your answer4 #ach4--,,the ,,artillerymen',s ,,dilemma4 ,illustration3 ,pyramid of cannon-balls ,m,m,m,m,m,m,mr ,,mm ,,mm3 ,m #j r,w,z,x ,m 3 ,,mwm a,x 1,,bm ,m #j,m ,m a,m,m,m,m#b,,mw #jb ,m,m,w,m,mr ,,zm4 ,m,m #h,,mm #g ,,xm#b ,,ms#b ,m4,,mmmwmmmm ,,mm ,m ,,mx i,m,m ,m#g,w #h 4 ,m r ,w ,m ,z2,m ,m #jr 2 ,m ,m ,m ,w #bb ,w ,m ,m ,m ,m4,m#b,,wmmmmz 2,,mm,x3#g,,mmmb2 ,,mmm ,,zm ,m3,,mm#j2h3 1,,ms ,ma #h ,m,m,m,m,m,m,mi r,m #b,,mmmmmm ,,mb ,m #g ,,xm1 13 ,,bmm3 r#g,s 41,,mm ,,mm ,,mb ,m i 1,m 1 #b 2 a,m,m,m,m,m,m,m,m,m ,,xm2 ,,mzm ,m 4 ,m #g ,m 4 ,z ,m ,m ,m#h ,m ,m ,s ,m 4#j ,m #h,,mm a,mi3 ,,mmmm#g,m 1#g 4i,m ,x ,m a,z ,m ,m #h 1,m,m,m,m,b,m,m,ma ,,smw #g,m1,,xz,,mm ,m #h,m ,m 4,m ,,mmmm,x ,m,mr ,ma ,,mmmmmmmmm ,m 4,,wm ,m ,,wm#g,,wmm 4,,wx ,,mzs ,m #h,m 3,,mmmwmmmm #h,x ,,mmmbmmm#g #g;a,m #b,,mm r1 #hr ,,zm#b ,mr#b a,m,m2 ,mai 3,,ms 3i,m ,zi,m ,,mx ,m ,m 4 ,m ,wr4,m,m,ma,b,m,m,m,b ,m ,m ,,mz4 1,m ,,mmz ,mr ,m ,m ,b#j ,z #b,s i,m ,s ,,xm #g ,,wmm ,,mm 4,m ,m ,m ,w ,m4 ,m ,m #j2,m#b,m2,,mmmm3 ,,ww#h;a,m,m ,m ,s ,m ,m ,m 3 ,ma,m,m,m,m,m,m ,,mm#j,w2,,mzm3 ,m i ,m ,m ,,mm ,m,m,m,z,m,b,za#j;ar ,b#bjr,m,m,m ,si i ,,bw ,,mmm#jb #g,,mm#j #b,,mmm ,,mmmmmmmm 4 ,m ,,smai,m,m#bj;,,bwm ,,xm #j,,zmmm3,,mmmmw2 r4 #j,,wmbm#b,xr,b3 4 ,all cannon-balls are to be piled in square pyramids1 was the order issued to the regiment4 ,this was done4 ,then came the further order1 ,all pyramids are to contain a square number of balls4 ,whereupon the trouble arose4 ,it can't be done1 said the major4 ,look at this pyramid1 for example2 there are sixteen balls at the base1 then nine1 then four1 then one at the top1 making thirty balls in all4 ,but there must be six more balls1 or five fewer1 to make a square number4 ,it must be done1 insisted the general4 ,all you have to do is to put the right number of balls in your pyramids4 ,i've got it6 said a lieutenant1 the mathematical genius of the regiment4 ,lay the balls out singly4 ,bosh6 exclaimed the general4 ,you can't pile one ball into a pyramid6 ,is it really possible to obey both orders8 #aci4--,,the ,,dutchmen',s ,,wives4 ,i wonder how many of my readers are acquainted with the puzzle of the ,dutchmen's ,wives--in which you have to determine the names of three men's wives1 or1 rather1 which wife belongs to each husband4 ,some thirty years ago it was going the rounds1 as something quite new1 but ,i recently discovered it in the ,ladies' ,diary for #agci-dj1 so it was clearly familiar to the fair sex over one hundred and seventy years ago4 ,how many of our mothers1 wives1 sisters1 daughters1 and aunts could solve the puzzle to-day8 ,a far greater proportion than then1 let us hope4 ,three ,dutchmen1 named ,hendrick1 ,elas1 and ,cornelius1 and their wives1 ,gurtrn1 ,katrn1 and ,anna1 purchase hogs4 ,each buys as many as he 7or she7 gives shillings for one4 ,each husband pays altogether three guineas more than his wife4 ,hendrick buys twenty-three more hogs than ,katrn1 and ,elas eleven more than ,gurtrn4 ,now1 what was the name of each man's wife8 ,illustration #adj4--;,,find ,,ada',s ,,surname4 ,this puzzle closely resembles the last one1 my remarks on the solution of which the reader may like to apply in another case4 ,it was recently submitted to a ,sydney evening newspaper that indulges in intellect sharpeners1 but was rejected with the remark that it is childish and that they only published problems capable of solution6 ,five ladies1 accompanied by their daughters1 bought cloth at the same shop4 ,each of the ten paid as many farthings per foot as she bought feet1 and each mother spent #hs4 #e;d4 more than her daughter4 ,mrs4 ,robinson spent #fs4 more than ,mrs4 ,evans1 who spent about a quarter as much as ,mrs4 ,jones4 ,mrs4 ,smith spent most of all4 ,mrs4 ,brown bought #ba yards more than ,bessie--one of the girls4 ,annie bought #af yards more than ,mary and spent #c1 #js4 #h;d4 more than ,emily4 ,the ,christian name of the other girl was ,ada4 ,now1 what was her surname8 #ada4--,,saturday ,,marketing4 ,here is an amusing little case of marketing which1 although it deals with a good many items of money1 leads up to a question of a totally different character4 ,four married couples went into their village on a recent ,saturday night to do a little marketing4 ,they had to be very economical1 for among them they only possessed forty shilling coins4 ,the fact is1 ,ann spent #as41 ,mary spent #bs41 ,jane spent #cs41 and ,kate spent #ds4 ,the men were rather more extravagant than their wives1 for ,ned ,smith spent as much as his wife1 ,tom ,brown twice as much as his wife1 ,bill ,jones three times as much as his wife1 and ,jack ,robinson four times as much as his wife4 ,on the way home somebody suggested that they should divide what coin they had left equally among them4 ,this was done1 and the puzzling question is simply this3 ,what was the surname of each woman8 ,can you pair off the four couples8 ,,geometrical ,,problems4 ,god geometrizes continually4 ,,plato4 ,there is no study1 said ,augustus de ,morgan1 which presents so simple a beginning as that of geometry2 there is none in which difficulties grow more rapidly as we proceed4 ,this will be found when the reader comes to consider the following puzzles1 though they are not arranged in strict order of difficulty4 ,and the fact that they have interested and given pleasure to man for untold ages is no doubt due in some measure to the appeal they make to the eye as well as to the brain4 ,sometimes an algebraical formula or theorem seems to give pleasure to the mathematician's eye1 but it is probably only an intellectual pleasure4 ,but there can be no doubt that in the case of certain geometrical problems1 notably dissection or superposition puzzles1 the sthetic faculty in man contributes to the delight4 ,for example1 there are probably few readers who will examine the various cuttings of the ,greek cross in the following pages without being in some degree stirred by a sense of beauty4 ,law and order in ,nature are always pleasing to contemplate1 but when they come under the very eye they seem to make a specially strong appeal4 ,even the person with no geometrical knowledge whatever is induced after the inspection of such things to exclaim1 ,how very pretty6 ,in fact1 ,i have known more than one person led on to a study of geometry by the fascination of cutting-out puzzles4 ,i have1 therefore1 thought it well to keep these dissection puzzles distinct from the geometrical problems on more general lines4 ,,dissection ,,puzzles4 ,take him and cut him out in little stars4 ,romeo and ,juliet1 iii4 #b4 ,puzzles have infinite variety1 but perhaps there is no class more ancient than dissection1 cutting-out1 or superposition puzzles4 ,they were certainly known to the ,chinese several thousand years before the ,christian era4 ,and they are just as fascinating to-day as they can have been at any period of their history4 ,it is supposed by those who have investigated the matter that the ancient ,chinese philosophers used these puzzles as a sort of kindergarten method of imparting the principles of geometry4 ,whether this was so or not1 it is certain that all good dissection puzzles 7for the nursery type of jig-saw puzzle1 which merely consists in cutting up a picture into pieces to be put together again1 is not worthy of serious consideration7 are really based on geometrical laws4 ,this statement need not1 however1 frighten off the novice1 for it means little more than this1 that geometry will give us the reason why1 if we are interested in knowing it1 though the solutions may often be discovered by any intelligent person after the exercise of patience1 ingenuity1 and common sagacity4 ,if we want to cut one plane figure into parts that by readjustment will form another figure1 the first thing is to find a way of doing it at all1 and then to discover how to do it in the fewest possible pieces4 ,often a dissection problem is quite easy apart from this limitation of pieces4 ,at the time of the publication in the ,weekly ,dispatch1 in #aijb1 of a method of cutting an equilateral triangle into four parts that will form a square 7see ,no4 #bf1 ,canterbury ,puzzles71 no geometrician would have had any difficulty in doing what is required in five pieces3 the whole point of the discovery lay in performing the little feat in four pieces only4 ,mere approximations in the case of these problems are valueless2 the solution must be geometrically exact1 or it is not a solution at all4 ,fallacies are cropping up now and again1 and ,i shall have occasion to refer to one or two of these4 ,they are interesting merely as fallacies4 ,but ,i want to say something on two little points that are always arising in cutting-out puzzles--the questions of hanging by a thread and turning over4 ,these points can best be illustrated by a puzzle that is frequently to be found in the old books1 but invariably with a false solution4 ,the puzzle is to cut the figure shown in ,fig4 #a into three pieces that will fit together and form a half-square triangle4 ,the answer that is invariably given is that shown in ,figs4 #a and #b4 ,now1 it is claimed that the four pieces marked ,c are really only one piece1 because they may be so cut that they are left hanging together by a mere thread4 ,but no serious puzzle lover will ever admit this4 ,if the cut is made so as to leave the four pieces joined in one1 then it cannot result in a perfectly exact solution4 ,if1 on the other hand1 the solution is to be exact1 then there will be four pieces--or six pieces in all4 ,it is1 therefore1 not a solution in three pieces4 ,illustration3 ,fig4 #a ,illustration3 ,fig4 #b ,if1 however1 the reader will look at the solution in ,figs4 #c and #d1 he will see that no such fault can be found with it4 ,there is no question whatever that there are three pieces1 and the solution is in this respect quite satisfactory4 ,but another question arises4 ,it will be found on inspection that the piece marked ,f1 in ,fig4 #c1 is turned over in ,fig4 #d--that is to say1 a different side has necessarily to be presented4 ,if the puzzle were merely to be cut out of cardboard or wood1 there might be no objection to this reversal1 but it is quite possible that the material would not admit of being reversed4 ,there might be a pattern1 a polish1 a difference of texture1 that prevents it4 ,but it is generally understood that in dissection puzzles you are allowed to turn pieces over unless it is distinctly stated that you may not do so4 ,and very often a puzzle is greatly improved by the added condition1 no piece may be turned over4 ,i have often made puzzles1 too1 in which the diagram has a small repeated pattern1 and the pieces have then so to be cut that not only is there no turning over1 but the pattern has to be matched1 which cannot be done if the pieces are turned round1 even with the proper side uppermost4 ,illustration3 ,fig4 #c ,illustration3 ,fig4 #d ,before presenting a varied series of cutting-out puzzles1 some very easy and others difficult1 ,i propose to consider one family alone--those problems involving what is known as the ,greek cross with the square4 ,this will exhibit a great variety of curious transpositions1 and1 by having the solutions as we go along1 the reader will be saved the trouble of perpetually turning to another part of the book1 and will have everything under his eye4 ,it is hoped that in this way the article may prove somewhat instructive to the novice and interesting to others4 ,,greek ,,cross ,,puzzles4 ,to fret thy soul with crosses4 ,,spenser4 ,but1 for my part1 it was ,greek to me4 ,julius ,csar1 i4 #b4 ,many people are accustomed to consider the cross as a wholly ,christian symbol4 ,this is erroneous3 it is of very great antiquity4 ,the ancient ,egyptians employed it as a sacred symbol1 and on ,greek sculptures we find representations of a cake 7the supposed real origin of our hot cross buns7 bearing a cross4 ,two such cakes were discovered at ,herculaneum4 ,cecrops offered to ,jupiter ,olympus a sacred cake or boun of this kind4 ,the cross and ball1 so frequently found on ,egyptian figures1 is a circle and the tau cross4 ,the circle signified the eternal preserver of the world1 and the ,t1 named from the ,greek letter tau1 is the monogram of ,thoth1 the ,egyptian ,mercury1 meaning wisdom4 ,this tau cross is also called by ,christians the cross of ,st4 ,anthony1 and is borne on a badge in the bishop's palace at ,exeter4 ,as for the ,greek or mundane cross1 the cross with four equal arms1 we are told by competent antiquaries that it was regarded by ancient occultists for thousands of years as a sign of the dual forces of ,nature--the male and female spirit of everything that was everlasting4 ,illustration3 ,fig4 #e4 ,the ,greek cross1 as shown in ,fig4 #e1 is formed by the assembling together of five equal squares4 ,we will start with what is known as the ,hindu problem1 supposed to be upwards of three thousand years old4 ,it appears in the seal of ,harvard ,college1 and is often given in old works as symbolical of mathematical science and exactitude4 ,cut the cross into five pieces to form a square4 ,figs4 #f and #g show how this is done4 ,it was not until the middle of the nineteenth century that we found that the cross might be transformed into a square in only four pieces4 ,figs4 #h and #i will show how to do it1 if we further require the four pieces to be all of the same size and shape4 ,this ,fig4 #i is remarkable because1 according to ,dr4 ,le ,plongeon and others1 as expounded in a work by ,professor ,wilson of the ,smithsonian ,institute1 here we have the great ,swastika1 or sign1 of good luck to you --the most ancient symbol of the human race of which there is any record4 ,professor ,wilson's work gives some four hundred illustrations of this curious sign as found in the ,aztec mounds of ,mexico1 the pyramids of ,egypt1 the ruins of ,troy1 and the ancient lore of ,india and ,china4 ,one might almost say there is a curious affinity between the ,greek cross and ,swastika6 ,if1 however1 we require that the four pieces shall be produced by only two clips of the scissors 7assuming the puzzle is in paper form71 then we must cut as in ,fig4 #aj to form ,fig4 #aa1 the first clip of the scissors being from a to b4 ,of course folding the paper1 or holding the pieces together after the first cut1 would not in this case be allowed4 ,but there is an infinite number of different ways of making the cuts to solve the puzzle in four pieces4 ,to this point ,i propose to return4 ,illustration3 ,fig4 #f ,illustration3 ,fig4 #g ,illustration3 ,fig4 #h ,illustration3 ,fig4 #i ,illustration3 ,fig4 #aj ,illustration3 ,fig4 #aa ,it will be seen that every one of these puzzles has its reverse puzzle--to cut a square into pieces to form a ,greek cross4 ,but as a square has not so many angles as the cross1 it is not always equally easy to discover the true directions of the cuts4 ,yet in the case of the examples given1 ,i will leave the reader to determine their direction for himself1 as they are rather obvious from the diagrams4 ,cut a square into five pieces that will form two separate ,greek crosses of different sizes4 ,this is quite an easy puzzle4 ,as will be seen in ,fig4 #ab1 we have only to divide our square into #be little squares and then cut as shown4 ,the cross ,a is cut out entire1 and the pieces ,b1 ,c1 ,d1 and ,e form the larger cross in ,fig4 #ac4 ,the reader may here like to cut the single piece1 ,b1 into four pieces all similar in shape to itself1 and form a cross with them in the manner shown in ,fig4 #ac4 ,i hardly need give the solution4 ,illustration3 ,,fig4 #ab4 ,illustration3 ,,fig4 #ac4 ,cut a square into five pieces that will form two separate ,greek crosses of exactly the same size4 ,this is more difficult4 ,we make the cuts as in ,fig4 #ad1 where the cross ,a comes out entire and the other four pieces form the cross in ,fig4 #ae4 ,the direction of the cuts is pretty obvious4 ,it will be seen that the sides of the square in ,fig4 #ad are marked off into six equal parts4 ,the sides of the cross are found by ruling lines from certain of these points to others4 ,illustration3 ,,fig4 #ad4 ,illustration3 ,,fig4 #ae4 ,i will now explain1 as ,i promised1 why a ,greek cross may be cut into four pieces in an infinite number of different ways to make a square4 ,draw a cross1 as in ,fig4 #af4 ,then draw on transparent paper the square shown in ,fig4 #ag1 taking care that the distance c to d is exactly the same as the distance a to b in the cross4 ,now place the transparent paper over the cross and slide it about into different positions1 only be very careful always to keep the square at the same angle to the cross as shown1 where a b is parallel to c d4 ,if you place the point c exactly over a the lines will indicate the solution 7,figs4 #aj and #aa74 ,if you place c in the very centre of the dotted square1 it will give the solution in ,figs4 #h and #i4 ,you will now see that by sliding the square about so that the point c is always within the dotted square you may get as many different solutions as you like2 because1 since an infinite number of different points may theoretically be placed within this square1 there must be an infinite number of different solutions4 ,but the point c need not necessarily be placed within the dotted square4 ,it may be placed1 for example1 at point e to give a solution in four pieces4 ,here the joins at a and f may be as slender as you like4 ,yet if you once get over the edge at a or f you no longer have a solution in four pieces4 ,this proof will be found both entertaining and instructive4 ,if you do not happen to have any transparent paper at hand1 any thin paper will of course do if you hold the two sheets against a pane of glass in the window4 ,illustration3 ,,fig4 #af4 ,illustration3 ,,fig4 #ag4 ,it may have been noticed from the solutions of the puzzles that ,i have given that the side of the square formed from the cross is always equal to the distance a to b in ,fig4 #af4 ,this must necessarily be so1 and ,i will presently try to make the point quite clear4 ,we will now go one step further4 ,i have already said that the ideal solution to a cutting-out puzzle is always that which requires the fewest possible pieces4 ,we have just seen that two crosses of the same size may be cut out of a square in five pieces4 ,the reader who succeeded in solving this perhaps asked himself3 ,can it be done in fewer pieces8 ,this is just the sort of question that the true puzzle lover is always asking1 and it is the right attitude for him to adopt4 ,the answer to the question is that the puzzle may be solved in four pieces--the fewest possible4 ,this1 then1 is a new puzzle4 ,cut a square into four pieces that will form two ,greek crosses of the same size4 ,illustration3 ,,fig4 #ah4 ,illustration3 ,,fig4 #ai4 ,illustration3 ,,fig4 #bj4 ,the solution is very beautiful4 ,if you divide by points the sides of the square into three equal parts1 the directions of the lines in ,fig4 #ah will be quite obvious4 ,if you cut along these lines1 the pieces ,a and ,b will form the cross in ,fig4 #ai and the pieces ,c and ,d the similar cross in ,fig4 #bj4 ,in this square we have another form of ,swastika4 ,the reader will here appreciate the truth of my remark to the effect that it is easier to find the directions of the cuts when transforming a cross to a square than when converting a square into a cross4 ,thus1 in ,figs4 #f1 #h1 and #aj the directions of the cuts are more obvious than in ,fig4 #ad1 where we had first to divide the sides of the square into six equal parts1 and in ,fig4 #ah1 where we divide them into three equal parts4 ,then1 supposing you were required to cut two equal ,greek crosses1 each into two pieces1 to form a square1 a glance at ,figs4 #ai and #bj will show how absurdly more easy this is than the reverse puzzle of cutting the square to make two crosses4 ,referring to my remarks on fallacies1 ,i will now give a little example of these solutions that are not solutions4 ,some years ago a young correspondent sent me what he evidently thought was a brilliant new discovery--the transforming of a square into a ,greek cross in four pieces by cuts all parallel to the sides of the square4 ,i give his attempt in ,figs4 #ba and #bb1 where it will be seen that the four pieces do not form a symmetrical ,greek cross1 because the four arms are not really squares but oblongs4 ,to make it a true ,greek cross we should require the additions that ,i have indicated with dotted lines4 ,of course his solution produces a cross1 but it is not the symmetrical ,greek variety required by the conditions of the puzzle4 ,my young friend thought his attempt was near enough to be correct2 but if he bought a penny apple with a sixpence he probably would not have thought it near enough if he had been given only fourpence change4 ,as the reader advances he will realize the importance of this question of exactitude4 ,illustration3 ,,fig4 #ba4 ,illustration3 ,,fig4 #bb4 ,in these cutting-out puzzles it is necessary not only to get the directions of the cutting lines as correct as possible1 but to remember that these lines have no width4 ,if after cutting up one of the crosses in a manner indicated in these articles you find that the pieces do not exactly fit to form a square1 you may be certain that the fault is entirely your own4 ,either your cross was not exactly drawn1 or your cuts were not made quite in the right directions1 or 7if you used wood and a fret-saw7 your saw was not sufficiently fine4 ,if you cut out the puzzles in paper with scissors1 or in cardboard with a penknife1 no material is lost2 but with a saw1 however fine1 there is a certain loss4 ,in the case of most puzzles this slight loss is not sufficient to be appreciable1 if the puzzle is cut out on a large scale1 but there have been instances where ,i have found it desirable to draw and cut out each part separately--not from one diagram--in order to produce a perfect result4 ,illustration3 ,,fig4 #bc4 ,illustration3 ,,fig4 #bd4 ,now for another puzzle4 ,if you have cut out the five pieces indicated in ,fig4 #ad1 you will find that these can be put together so as to form the curious cross shown in ,fig4 #bc4 ,so if ,i asked you to cut ,fig4 #bd into five pieces to form either a square or two equal ,greek crosses you would know how to do it4 ,you would make the cuts as in ,fig4 #bc1 and place them together as in ,figs4 #ad and #ae4 ,but ,i want something better than that1 and it is this4 ,cut ,fig4 #bd into only four pieces that will fit together and form a square4 ,illustration3 ,,fig4 #be4 ,illustration3 ,,fig4 #bf4 ,the solution to the puzzle is shown in ,figs4 #be and #bf4 ,the direction of the cut dividing ,a and ,c in the first diagram is very obvious1 and the second cut is made at right angles to it4 ,that the four pieces should fit together and form a square will surprise the novice1 who will do well to study the puzzle with some care1 as it is most instructive4 ,i will now explain the beautiful rule by which we determine the size of a square that shall have the same area as a ,greek cross1 for it is applicable1 and necessary1 to the solution of almost every dissection puzzle that we meet with4 ,it was first discovered by the philosopher ,pythagoras1 who died #ejj ,b4,c41 and is the #dgth proposition of ,euclid4 ,the young reader who knows nothing of the elements of geometry will get some idea of the fascinating character of that science4 ,the triangle ,,abc in ,fig4 #bg is what we call a right-angled triangle1 because the side ,,bc is at right angles to the side ,,ab4 ,now if we build up a square on each side of the triangle1 the squares on ,,ab and ,,bc will together be exactly equal to the square on the long side ,,ac1 which we call the hypotenuse4 ,this is proved in the case ,i have given by subdividing the three squares into cells of equal dimensions4 ,illustration3 ,,fig4 #bg4 ,illustration3 ,,fig4 #bh4 ,it will be seen that #i added to #af equals #be1 the number of cells in the large square4 ,if you make triangles with the sides #e1 #ab and #ac1 or with #h1 #ae and #ag1 you will get similar arithmetical proofs1 for these are all rational right-angled triangles1 but the law is equally true for all cases4 ,supposing we cut off the lower arm of a ,greek cross and place it to the left of the upper arm1 as in ,fig4 #bh1 then the square on ,,ef added to the square on ,,de exactly equals a square on ,,df4 ,therefore we know that the square of ,,df will contain the same area as the cross4 ,this fact we have proved practically by the solutions of the earlier puzzles of this series4 ,but whatever length we give to ,,de and ,,ef1 we can never give the exact length of ,,df in numbers1 because the triangle is not a rational one4 ,but the law is none the less geometrically true4 ,illustration3 ,,fig4 #bi4 ,illustration3 ,,fig4 #cj4 ,now look at ,fig4 #bi1 and you will see an elegant method for cutting a piece of wood of the shape of two squares 7of any relative dimensions7 into three pieces that will fit together and form a single square4 ,if you mark off the distance ab equal to the side cd the directions of the cuts are very evident4 ,from what we have just been considering1 you will at once see why bc must be the length of the side of the new square4 ,make the experiment as often as you like1 taking different relative proportions for the two squares1 and you will find the rule always come true4 ,if you make the two squares of exactly the same size1 you will see that the diagonal of any square is always the side of a square that is twice the size4 ,all this1 which is so simple that anybody can understand it1 is very essential to the solving of cutting-out puzzles4 ,it is in fact the key to most of them4 ,and it is all so beautiful that it seems a pity that it should not be familiar to everybody4 ,we will now go one step further and deal with the half-square4 ,take a square and cut it in half diagonally4 ,now try to discover how to cut this triangle into four pieces that will form a ,greek cross4 ,the solution is shown in ,figs4 #ca and #cb4 ,in this case it will be seen that we divide two of the sides of the triangle into three equal parts and the long side into four equal parts4 ,then the direction of the cuts will be easily found4 ,it is a pretty puzzle1 and a little more difficult than some of the others that ,i have given4 ,it should be noted again that it would have been much easier to locate the cuts in the reverse puzzle of cutting the cross to form a half-square triangle4 ,illustration3 ,,fig4 #ca4 ,illustration3 ,,fig4 #cb4 ,illustration3 ,,fig4 #cc4 ,illustration3 ,,fig4 #cd4 ,another ideal that the puzzle maker always keeps in mind is to contrive that there shall1 if possible1 be only one correct solution4 ,thus1 in the case of the first puzzle1 if we only require that a ,greek cross shall be cut into four pieces to form a square1 there is1 as ,i have shown1 an infinite number of different solutions4 ,it makes a better puzzle to add the condition that all the four pieces shall be of the same size and shape1 because it can then be solved in only one way1 as in ,figs4 #h and #i4 ,in this way1 too1 a puzzle that is too easy to be interesting may be improved by such an addition4 ,let us take an example4 ,we have seen in ,fig4 #bh that ,fig4 #cc can be cut into two pieces to form a ,greek cross4 ,i suppose an intelligent child would do it in five minutes4 ,but suppose we say that the puzzle has to be solved with a piece of wood that has a bad knot in the position shown in ,fig4 #cc--;a knot that we must not attempt to cut through--then a solution in two pieces is barred out1 and it becomes a more interesting puzzle to solve it in three pieces4 ,i have shown in ,figs4 #cc and #cd one way of doing this1 and it will be found entertaining to discover other ways of doing it4 ,of course ,i could bar out all these other ways by introducing more knots1 and so reduce the puzzle to a single solution1 but it would then be overloaded with conditions4 ,and this brings us to another point in seeking the ideal4 ,do not overload your conditions1 or you will make your puzzle too complex to be interesting4 ,the simpler the conditions of a puzzle are1 the better4 ,the solution may be as complex and difficult as you like1 or as happens1 but the conditions ought to be easily understood1 or people will not attempt a solution4 ,if the reader were now asked to cut a half-square into as few pieces as possible to form a ,greek cross1 he would probably produce our solution1 ,figs4 #ca-cb1 and confidently claim that he had solved the puzzle correctly4 ,in this way he would be wrong1 because it is not now stated that the square is to be divided diagonally4 ,although we should always observe the exact conditions of a puzzle we must not read into it conditions that are not there4 ,many puzzles are based entirely on the tendency that people have to do this4 ,the very first essential in solving a puzzle is to be sure that you understand the exact conditions4 ,now1 if you divided your square in half so as to produce ,fig4 #ce it is possible to cut it into as few as three pieces to form a ,greek cross4 ,we thus save a piece4 ,i give another puzzle in ,fig4 #cf4 ,the dotted lines are added merely to show the correct proportions of the figure--a square of #be cells with the four corner cells cut out4 ,the puzzle is to cut this figure into five pieces that will form a ,greek cross 7entire7 and a square4 ,illustration3 ,,fig4 #ce4 ,illustration3 ,,fig4 #cf4 ,the solution to the first of the two puzzles last given--to cut a rectangle of the shape of a half-square into three pieces that will form a ,greek cross--is shown in ,figs4 #cg and #ch4 ,it will be seen that we divide the long sides of the oblong into six equal parts and the short sides into three equal parts1 in order to get the points that will indicate the direction of the cuts4 ,the reader should compare this solution with some of the previous illustrations4 ,he will see1 for example1 that if we continue the cut that divides ,b and ,c in the cross1 we get ,fig4 #ae4 ,illustration3 ,,fig4 #cg4 ,illustration3 ,,fig4 #ch4 ,the other puzzle1 like the one illustrated in ,figs4 #ab and #ac1 will show how useful a little arithmetic may sometimes prove to be in the solution of dissection puzzles4 ,there are twenty-one of those little square cells into which our figure is subdivided1 from which we have to form both a square and a ,greek cross4 ,now1 as the cross is built up of five squares1 and #e from #ba leaves #af--;a square number--we ought easily to be led to the solution shown in ,fig4 #ci4 ,it will be seen that the cross is cut out entire1 while the four remaining pieces form the square in ,fig4 #dj4 ,illustration3 ,,fig4 #ci ,illustration3 ,,fig4 #dj ,of course a half-square rectangle is the same as a double square1 or two equal squares joined together4 ,therefore1 if you want to solve the puzzle of cutting a ,greek cross into four pieces to form two separate squares of the same size1 all you have to do is to continue the short cut in ,fig4 #ch right across the cross1 and you will have four pieces of the same size and shape4 ,now divide ,fig4 #cg into two equal squares by a horizontal cut midway and you will see the four pieces forming the two squares4 ,illustration3 ,,fig4 #da ,cut a ,greek cross into five pieces that will form two separate squares1 one of which shall contain half the area of one of the arms of the cross4 ,in further illustration of what ,i have already written1 if the two squares of the same size ,a ,b ,c ,d and ,b ,c ,f ,e1 in ,fig4 #da1 are cut in the manner indicated by the dotted lines1 the four pieces will form the large square ,a ,g ,e ,c4 ,we thus see that the diagonal ,a ,c is the side of a square twice the size of ,a ,b ,c ,d4 ,it is also clear that half the diagonal of any square is equal to the side of a square of half the area4 ,therefore1 if the large square in the diagram is one of the arms of your cross1 the small square is the size of one of the squares required in the puzzle4 ,the solution is shown in ,figs4 #db and #dc4 ,it will be seen that the small square is cut out whole and the large square composed of the four pieces ,b1 ,c1 ,d1 and ,e4 ,after what ,i have written1 the reader will have no difficulty in seeing that the square ,a is half the size of one of the arms of the cross1 because the length of the diagonal of the former is clearly the same as the side of the latter4 ,the thing is now self-evident4 ,i have thus tried to show that some of these puzzles that many people are apt to regard as quite wonderful and bewildering1 are really not difficult if only we use a little thought and judgment4 ,in conclusion of this particular subject ,i will give four ,greek cross puzzles1 with detached solutions4 #adb4--,,the ,,silk ,,patchwork4 ,the lady members of the ,wilkinson family had made a simple patchwork quilt1 as a small ,christmas present1 all composed of square pieces of the same size1 as shown in the illustration4 ,it only lacked the four corner pieces to make it complete4 ,somebody pointed out to them that if you unpicked the ,greek cross in the middle and then cut the stitches along the dark joins1 the four pieces all of the same size and shape would fit together and form a square4 ,this the reader knows1 from the solution in ,fig4 #ci1 is quite easily done4 ,but ,george ,wilkinson suddenly suggested to them this poser4 ,he said1 ,instead of picking out the cross entire1 and forming the square from four equal pieces1 can you cut out a square entire and four equal pieces that will form a perfect ,greek cross8 ,the puzzle is1 of course1 now quite easy4 #adc4--,,two ,,crosses ,,from ,,one4 ,cut a ,greek cross into five pieces that will form two such crosses1 both of the same size4 ,the solution of this puzzle is very beautiful4 #add4--,,the ,,cross ,,and ,,the ,,triangle4 ,cut a ,greek cross into six pieces that will form an equilateral triangle4 ,this is another hard problem1 and ,i will state here that a solution is practically impossible without a previous knowledge of my method of transforming an equilateral triangle into a square 7see ,no4 #bf1 ,canterbury ,puzzles74 #ade4--,,the ,,folded ,,cross4 ,cut out of paper a ,greek cross2 then so fold it that with a single straight cut of the scissors the four pieces produced will form a square4 ,,various ,,dissection ,,puzzles4 ,we will now consider a small miscellaneous selection of cutting-out puzzles1 varying in degrees of difficulty4 #adf4--;,,an ,,easy ,,dissection ,,puzzle4 ,first1 cut out a piece of paper or cardboard of the shape shown in the illustration4 ,it will be seen at once that the proportions are simply those of a square attached to half of another similar square1 divided diagonally4 ,the puzzle is to cut it into four pieces all of precisely the same size and shape4 #adg4--;,,an ,,easy ,,square ,,puzzle4 ,if you take a rectangular piece of cardboard1 twice as long as it is broad1 and cut it in half diagonally1 you will get two of the pieces shown in the illustration4 ,the puzzle is with five such pieces of equal size to form a square4 ,one of the pieces may be cut in two1 but the others must be used intact4 #adh4--,,the ,,bun ,,puzzle4 ,,the three circles represent three buns1 and it is simply required to show how these may be equally divided among four boys4 ,the buns must be regarded as of equal thickness throughout and of equal thickness to each other4 ,of course1 they must be cut into as few pieces as possible4 ,to simplify it ,i will state the rather surprising fact that only five pieces are necessary1 from which it will be seen that one boy gets his share in two pieces and the other three receive theirs in a single piece4 ,i am aware that this statement gives away the puzzle1 but it should not destroy its interest to those who like to discover the reason why4 #adi4--,,the ,,chocolate ,,squares4 ,here is a slab of chocolate1 indented at the dotted lines so that the twenty squares can be easily separated4 ,make a copy of the slab in paper or cardboard and then try to cut it into nine pieces so that they will form four perfect squares all of exactly the same size4 #aej4--;,,dissecting ,a ,,mitre4 ,the figure that is perplexing the carpenter in the illustration represents a mitre4 ,it will be seen that its proportions are those of a square with one quarter removed4 ,the puzzle is to cut it into five pieces that will fit together and form a perfect square4 ,i show an attempt1 published in ,america1 to perform the feat in four pieces1 based on what is known as the step principle1 but it is a fallacy4 ,illustration ,we are told first to cut oft the pieces #a and #b and pack them into the triangular space marked off by the dotted line1 and so form a rectangle4 ,so far1 so good4 ,now1 we are directed to apply the old step principle1 as shown1 and1 by moving down the piece #d one step1 form the required square4 ,but1 unfortunately1 it does not produce a square3 only an oblong4 ,call the three long sides of the mitre #hd in4 each4 ,then1 before cutting the steps1 our rectangle in three pieces will be #hd #fc4 ,the steps must be #aj in4 in height and #ab in4 in breadth4 ,therefore1 by moving down a step we reduce by #ab in4 the side #hd in4 and increase by #aj in4 the side #fc in4 ,hence our final rectangle must be #gb in4 #gc in41 which certainly is not a square6 ,the fact is1 the step principle can only be applied to rectangles with sides of particular relative lengths4 ,for example1 if the shorter side in this case were #fae_/g 7instead of #fc71 then the step method would apply4 ,for the steps would then be #ajb_/g in4 in height and #ab in4 in breadth4 ,note that #fae_/g #hd the square of #gb4 ,at present no solution has been found in four pieces1 and ,i do not believe one possible4 #aea4--,,the ,,joiner',s ,,problem4 ,i have often had occasion to remark on the practical utility of puzzles1 arising out of an application to the ordinary affairs of life of the little tricks and wrinkles that we learn while solving recreation problems4 ,illustration ,the joiner1 in the illustration1 wants to cut the piece of wood into as few pieces as possible to form a square table-top1 without any waste of material4 ,how should he go to work8 ,how many pieces would you require8 #aeb4--;,,another ,,joiner',s ,,problem4 ,illustration ,a joiner had two pieces of wood of the shapes and relative proportions shown in the diagram4 ,he wished to cut them into as few pieces as possible so that they could be fitted together1 without waste1 to form a perfectly square table-top4 ,how should he have done it8 ,there is no necessity to give measurements1 for if the smaller piece 7which is half a square7 be made a little too large or a little too small it will not affect the method of solution4 #aec--;,a ,,cutting-,,out ,,puzzle4 ,here is a little cutting-out poser4 ,i take a strip of paper1 measuring five inches by one inch1 and1 by cutting it into five pieces1 the parts fit together and form a square1 as shown in the illustration4 ,now1 it is quite an interesting puzzle to discover how we can do this in only four pieces4 ,illustration #aed4--,,mrs4 ,,hobson',s ,,hearthrug4 ,illustration ,mrs4 ,hobson's boy had an accident when playing with the fire1 and burnt two of the corners of a pretty hearthrug4 ,the damaged corners have been cut away1 and it now has the appearance and proportions shown in my diagram4 ,how is ,mrs4 ,hobson to cut the rug into the fewest possible pieces that will fit together and form a perfectly square rug8 ,it will be seen that the rug is in the proportions #cf #bg 7it does not matter whether we say inches or yards71 and each piece cut away measured #ab and #f on the outside4 #aee4--,,the ,,pentagon ,,and ,,square4 ,i wonder how many of my readers1 amongst those who have not given any close attention to the elements of geometry1 could draw a regular pentagon1 or five-sided figure1 if they suddenly required to do so4 ,a regular hexagon1 or six-sided figure1 is easy enough1 for everybody knows that all you have to do is to describe a circle and then1 taking the radius as the length of one of the sides1 mark off the six points round the circumference4 ,but a pentagon is quite another matter4 ,so1 as my puzzle has to do with the cutting up of a regular pentagon1 it will perhaps be well if ,i first show my less experienced readers how this figure is to be correctly drawn4 ,describe a circle and draw the two lines ,h ,b and ,d ,g1 in the diagram1 through the centre at right angles4 ,now find the point ,a1 midway between ,c and ,b4 ,next place the point of your compasses at ,a and with the distance ,a ,d describe the arc cutting ,h ,b at ,e4 ,then place the point of your compasses at ,d and with the distance ,d ,e describe the arc cutting the circumference at ,f4 ,now1 ,d ,f is one of the sides of your pentagon1 and you have simply to mark off the other sides round the circle4 ,quite simple when you know how1 but otherwise somewhat of a poser4 ,illustration ,having formed your pentagon1 the puzzle is to cut it into the fewest possible pieces that will fit together and form a perfect square4 ,illustration #aef4--,,the ,,dissected ,,triangle4 ,a good puzzle is that which the gentleman in the illustration is showing to his friends4 ,he has simply cut out of paper an equilateral triangle--that is1 a triangle with all its three sides of the same length4 ,he proposes that it shall be cut into five pieces in such a way that they will fit together and form either two or three smaller equilateral triangles1 using all the material in each case4 ,can you discover how the cuts should be made8 ,remember that when you have made your five pieces1 you must be able1 as desired1 to put them together to form either the single original triangle or to form two triangles or to form three triangles--all equilateral4 #aeg4--,,the ,,table-,,top ,,and ,,stools4 ,i have frequently had occasion to show that the published answers to a great many of the oldest and most widely known puzzles are either quite incorrect or capable of improvement4 ,i propose to consider the old poser of the table-top and stools that most of my readers have probably seen in some form or another in books compiled for the recreation of childhood4 ,the story is told that an economical and ingenious schoolmaster once wished to convert a circular table-top1 for which he had no use1 into seats for two oval stools1 each with a hand-hole in the centre4 ,he instructed the carpenter to make the cuts as in the illustration and then join the eight pieces together in the manner shown4 ,so impressed was he with the ingenuity of his performance that he set the puzzle to his geometry class as a little study in dissection4 ,but the remainder of the story has never been published1 because1 so it is said1 it was a characteristic of the principals of academies that they would never admit that they could err4 ,i get my information from a descendant of the original boy who had most reason to be interested in the matter4 ,the clever youth suggested modestly to the master that the hand-holes were too big1 and that a small boy might perhaps fall through them4 ,he therefore proposed another way of making the cuts that would get over this objection4 ,for his impertinence he received such severe chastisement that he became convinced that the larger the hand-hole in the stools the more comfortable might they be4 ,illustration ,now what was the method the boy proposed8 ,can you show how the circular table-top may be cut into eight pieces that will fit together and form two oval seats for stools 7each of exactly the same size and shape7 and each having similar hand-holes of smaller dimensions than in the case shown above8 ,of course1 all the wood must be used4 #aeh4--,,the ,,great ,,monad4 ,illustration ,here is a symbol of tremendous antiquity which is worthy of notice4 ,it is borne on the ,korean ensign and merchant flag1 and has been adopted as a trade sign by the ,northern ,pacific ,railroad ,company1 though probably few are aware that it is the ,great ,monad1 as shown in the sketch below4 ,this sign is to the ,chinaman what the cross is to the ,christian4 ,it is the sign of ,deity and eternity1 while the two parts into which the circle is divided are called the ,yin and the ,yan--the male and female forces of nature4 ,a writer on the subject more than three thousand years ago is reported to have said in reference to it3 ,the illimitable produces the great extreme4 ,the great extreme produces the two principles4 ,the two principles produce the four quarters1 and from the four quarters we develop the quadrature of the eight diagrams of ,feuh-hi4 ,i hope readers will not ask me to explain this1 for ,i have not the slightest idea what it means4 ,yet ,i am persuaded that for ages the symbol has had occult and probably mathematical meanings for the esoteric student4 ,i will introduce the ,monad in its elementary form4 ,here are three easy questions respecting this great symbol3-- 7,i47 ,which has the greater area1 the inner circle containing the ,yin and the ,yan1 or the outer ring8 7,,ii47 ,divide the ,yin and the ,yan into four pieces of the same size and shape by one cut4 7,,iii47 ,divide the ,yin and the ,yan into four pieces of the same size1 but different shape1 by one straight cut4 #aei4--,,the ,,square ,,of ,,veneer4 ,the following represents a piece of wood in my possession1 #e in4 square4 ,by markings on the surface it is divided into twenty-five square inches4 ,i want to discover a way of cutting this piece of wood into the fewest possible pieces that will fit together and form two perfect squares of different sizes and of known dimensions4 ,but1 unfortunately1 at every one of the sixteen intersections of the cross lines a small nail has been driven in at some time or other1 and my fret-saw will be injured if it comes in contact with any of these4 ,i have therefore to find a method of doing the work that will not necessitate my cutting through any of those sixteen points4 ,how is it to be done8 ,remember1 the exact dimensions of the two squares must be given4 ,illustration #afj4--,,the ,,two ,,horseshoes4 ,illustration ,why horseshoes should be considered lucky is one of those things which no man can understand4 ,it is a very old superstition1 and ,john ,aubrey 7#afbf-agjj7 says1 ,most houses at the ,west ,end of ,london have a horseshoe on the threshold4 ,in ,monmouth ,street there were seventeen in #ahac and seven so late as #ahee4 ,even ,lord ,nelson had one nailed to the mast of the ship ,victory4 ,to-day we find it more conducive to good luck to see that they are securely nailed on the feet of the horse we are about to drive4 ,nevertheless1 so far as the horseshoe1 like the ,swastika and other emblems that ,i have had occasion at times to deal with1 has served to symbolize health1 prosperity1 and goodwill towards men1 we may well treat it with a certain amount of respectful interest4 ,may there not1 moreover1 be some esoteric or lost mathematical mystery concealed in the form of a horseshoe8 ,i have been looking into this matter1 and ,i wish to draw my readers' attention to the very remarkable fact that the pair of horseshoes shown in my illustration are related in a striking and beautiful manner to the circle1 which is the symbol of eternity4 ,i present this fact in the form of a simple problem1 so that it may be seen how subtly this relation has been concealed for ages and ages4 ,my readers will1 ,i know1 be pleased when they find the key to the mystery4 ,cut out the two horseshoes carefully round the outline and then cut them into four pieces1 all different in shape1 that will fit together and form a perfect circle4 ,each shoe must be cut into two pieces and all the part of the horse's hoof contained within the outline is to be used and regarded as part of the area4 #afa4--,,the ,,betsy ,,ross ,,puzzle4 ,a correspondent asked me to supply him with the solution to an old puzzle that is attributed to a certain ,betsy ,ross1 of ,philadelphia1 who showed it to ,george ,washington4 ,it consists in so folding a piece of paper that with one clip of the scissors a five-pointed star of ,freedom may be produced4 ,whether the story of the puzzle's origin is a true one or not ,i cannot say1 but ,i have a print of the old house in ,philadelphia where the lady is said to have lived1 and ,i believe it still stands there4 ,but my readers will doubtless be interested in the little poser4 ,take a circular piece of paper and so fold it that with one cut of the scissors you can produce a perfect five-pointed star4 #afb4--,,the ,,cardboard ,,chain4 ,illustration ,can you cut this chain out of a piece of cardboard without any join whatever8 ,every link is solid2 without its having been split and afterwards joined at any place4 ,it is an interesting old puzzle that ,i learnt as a child1 but ,i have no knowledge as to its inventor4 #afc4--,,the ,,paper ,,box4 ,it may be interesting to introduce here1 though it is not strictly a puzzle1 an ingenious method for making a paper box4 ,take a square of stout paper and by successive foldings make all the creases indicated by the dotted lines in the illustration4 ,then cut away the eight little triangular pieces that are shaded1 and cut through the paper along the dark lines4 ,the second illustration shows the box half folded up1 and the reader will have no difficulty in effecting its completion4 ,before folding up1 the reader might cut out the circular piece indicated in the diagram1 for a purpose ,i will now explain4 ,this box will be found to serve excellently for the production of vortex rings4 ,these rings1 which were discussed by ,von ,helmholtz in #aheh1 are most interesting1 and the box 7with the hole cut out7 will produce them to perfection4 ,fill the box with tobacco smoke by blowing it gently through the hole4 ,now1 if you hold it horizontally1 and softly tap the side that is opposite to the hole1 an immense number of perfect rings can be produced from one mouthful of smoke4 ,it is best that there should be no currents of air in the room4 ,people often do not realise that these rings are formed in the air when no smoke is used4 ,the smoke only makes them visible4 ,now1 one of these rings1 if properly directed on its course1 will travel across the room and put out the flame of a candle1 and this feat is much more striking if you can manage to do it without the smoke4 ,of course1 with a little practice1 the rings may be blown from the mouth1 but the box produces them in much greater perfection1 and no skill whatever is required4 ,lord ,kelvin propounded the theory that matter may consist of vortex rings in a fluid that fills all space1 and by a development of the hypothesis he was able to explain chemical combination4 ,illustration3 ---------------------------------------- ---- _/ _/ ---------------------------------------- ---- ,illustration #afd4--,,the ,,potato ,,puzzle4 ,take a circular slice of potato1 place it on the table1 and see into how large a number of pieces you can divide it with six cuts of a knife4 ,of course you must not readjust the pieces or pile them after a cut4 ,what is the greatest number of pieces you can make8 ,illustration3 -------- _/ #a_/ _/ #b _/ #c _/ _/ _/ _/ _/ #d _/ #e_/ #f _/ _/ _/ _/ #g_/ #h_/ #i_/aj _/ _/ _/ _/ _/ _/#aa_/ab_/ac_/ _/ _/ _/ _/#ad_/ae_/ _/ _/ _/#af_/ ----- ,the illustration shows how to make sixteen pieces4 ,this can1 of course1 be easily beaten4 #afe4--,,the ,,seven ,,pigs4 ,illustration ------------------------------ ,p ,p ,p ,p ,p ,p ,p ------------------------------ ,here is a little puzzle that was put to one of the sons of ,erin the other day and perplexed him unduly1 for it is really quite easy4 ,it will be seen from the illustration that he was shown a sketch of a square pen containing seven pigs4 ,he was asked how he would intersect the pen with three straight fences so as to enclose every pig in a separate sty4 ,in other words1 all you have to do is to take your pencil and1 with three straight strokes across the square1 enclose each pig separately4 ,nothing could be simpler4 ,illustration ,the ,irishman complained that the pigs would not keep still while he was putting up the fences4 ,he said that they would all flock together1 or one obstinate beast would go into a corner and flock all by himself4 ,it was pointed out to him that for the purposes of the puzzle the pigs were stationary4 ,he answered that ,irish pigs are not stationery--they are pork4 ,being persuaded to make the attempt1 he drew three lines1 one of which cut through a pig4 ,when it was explained that this is not allowed1 he protested that a pig was no use until you cut its throat4 ,begorra1 if it's bacon ye want without cutting your pig1 it will be all gammon4 ,we will not do the ,irishman the injustice of suggesting that the miserable pun was intentional4 ,however1 he failed to solve the puzzle4 ,can you do it8 #aff4--,,the ,,landowner',s ,,fences4 ,the landowner in the illustration is consulting with his bailiff over a rather puzzling little question4 ,he has a large plan of one of his fields1 in which there are eleven trees4 ,now1 he wants to divide the field into just eleven enclosures by means of straight fences1 so that every enclosure shall contain one tree as a shelter for his cattle4 ,how is he to do it with as few fences as possible8 ,take your pencil and draw straight lines across the field until you have marked off the eleven enclosures 7and no more71 and then see how many fences you require4 ,of course the fences may cross one another4 #afg4--,,the ,,wizard',s ,,cats4 ,illustration ,a wizard placed ten cats inside a magic circle as shown in our illustration1 and hypnotized them so that they should remain stationary during his pleasure4 ,he then proposed to draw three circles inside the large one1 so that no cat could approach another cat without crossing a magic circle4 ,try to draw the three circles so that every cat has its own enclosure and cannot reach another cat without crossing a line4 #afh4--,,the ,,christmas ,,pudding4 ,illustration ,speaking of ,christmas puddings1 said the host1 as he glanced at the imposing delicacy at the other end of the table4 ,i am reminded of the fact that a friend gave me a new puzzle the other day respecting one4 ,here it is1 he added1 diving into his breast pocket4 ',problem3 ,to find the contents1' ,i suppose1 said the ,eton boy4 ,no2 the proof of that is in the eating4 ,i will read you the conditions4 ',cut the pudding into two parts1 each of exactly the same size and shape1 without touching any of the plums4 ,the pudding is to be regarded as a flat disc1 not as a sphere4' ,why should you regard a ,christmas pudding as a disc8 ,and why should any reasonable person ever wish to make such an accurate division8 asked the cynic4 ,it is just a puzzle--a problem in dissection4 ,all in turn had a look at the puzzle1 but nobody succeeded in solving it4 ,it is a little difficult unless you are acquainted with the principle involved in the making of such puddings1 but easy enough when you know how it is done4 #afi4--;,a ,,tangram ,,paradox4 ,many pastimes of great antiquity1 such as chess1 have so developed and changed down the centuries that their original inventors would scarcely recognize them4 ,this is not the case with ,tangrams1 a recreation that appears to be at least four thousand years old1 that has apparently never been dormant1 and that has not been altered or improved upon since the legendary ,chinaman ,tan first cut out the seven pieces shown in ,diagram ,i4 ,if you mark the point ,b1 midway between ,a and ,c1 on one side of a square of any size1 and ,d1 midway between ,c and ,e1 on an adjoining side1 the direction of the cuts is too obvious to need further explanation4 ,every design in this article is built up from the seven pieces of blackened cardboard4 ,it will at once be understood that the possible combinations are infinite4 ,illustration ,the late ,mr4 ,sam ,loyd1 of ,new ,york1 who published a small book of very ingenious designs1 possessed the manuscripts of the late ,mr4 ,challenor1 who made a long and close study of ,tangrams4 ,this gentleman1 it is said1 records that there were originally seven books of ,tangrams1 compiled in ,china two thousand years before the ,christian era4 ,these books are so rare that1 after forty years' residence in the country1 he only succeeded in seeing perfect copies of the first and seventh volumes with fragments of the second4 ,portions of one of the books1 printed in gold leaf upon parchment1 were found in ,peking by an ,english soldier and sold for three hundred pounds4 ,a few years ago a little book came into my possession1 from the library of the late ,lewis ,carroll1 entitled ,the ,fashionable ,chinese ,puzzle4 ,it contains three hundred and twenty-three ,tangram designs1 mostly nondescript geometrical figures1 to be constructed from the seven pieces4 ,it was ,published by ,j4 and ,e4 ,wallis1 #db ,skinner ,street1 and ,j4 ,wallis1 ,jun41 ,marine ,library1 ,sidmouth 7,south ,devon74 ,there is no date1 but the following note fixes the time of publication pretty closely3 ,this ingenious contrivance has for some time past been the favourite amusement of the ex-,emperor ,napoleon1 who1 being now in a debilitated state and living very retired1 passes many hours a day in thus exercising his patience and ingenuity4 ,the reader will find1 as did the great exile1 that much amusement1 not wholly uninstructive1 may be derived from forming the designs of others4 ,he will find many of the illustrations to this article quite easy to build up1 and some rather difficult4 ,every picture may thus be regarded as a puzzle4 ,but it is another pastime altogether to create new and original designs of a pictorial character1 and it is surprising what extraordinary scope the ,tangrams afford for producing pictures of real life--angular and often grotesque1 it is true1 but full of character4 ,i give an example of a recumbent figure 7#b7 that is particularly graceful1 and only needs some slight reduction of its angularities to produce an entirely satisfactory outline4 ,as ,i have referred to the author of ,alice in ,wonderland1 ,i give also my designs of the ,march ,hare 7#c7 and the ,hatter 7#d74 ,i also give an attempt at ,napoleon 7#e71 and a very excellent ,red ,indian with his ,squaw by ,mr4 ,loyd 7#f and #g74 ,a large number of other designs will be found in an article by me in ,the ,strand ,magazine for ,november1 #aijh4 ,illustration3 #b ,illustration3 #c ,illustration3 #d ,on the appearance of this magazine article1 the late ,sir ,james ,murray1 the eminent philologist1 tried1 with that amazing industry that characterized all his work1 to trace the word tangram to its source4 ,at length he wrote as follows3--,one of my sons is a professor in the ,anglo-,chinese college at ,tientsin4 ,through him1 his colleagues1 and his students1 ,i was able to make inquiries as to the alleged ,tan among ,chinese scholars4 ,our ,chinese professor here 7,oxford7 also took an interest in the matter and obtained information from the secretary of the ,chinese ,legation in ,london1 who is a very eminent representative of the ,chinese literati4 ,illustration3 #e ,the result has been to show that the man ,tan1 the god ,tan1 and the ',book of ,tan' are entirely unknown to ,chinese literature1 history1 or tradition4 ,by most of the learned men the name1 or allegation of the existence1 of these had never been heard of4 ,the puzzle is1 of course1 well known4 ,it is called in ,chinese ch'i ch'iao t'u2 literally1 'seven-ingenious-plan' or 'ingenious-puzzle figure of seven pieces4' ,no name approaching 'tangram1' or even 'tan1' occurs in ,chinese1 and the only suggestions for the latter were the ,chinese t'an1 'to extend'2 or t'ang1 ,cantonese dialect for ',chinese4' ,it was suggested that probably some ,american or ,englishman who knew a little ,chinese or ,cantonese1 wanting a name for the puzzle1 might concoct one out of one of these words and the ,european ending 'gram4' ,i should say the name 'tangram' was probably invented by an ,american some little time before #ahfd and after #ahdg1 but ,i cannot find it in print before the #ahfd edition of ,webster4 ,i have therefore had to deal very shortly with the word in the dictionary1 telling what it is applied to and what conjectures or guesses have been made at the name1 and giving a few quotations1 one from your own article1 which has enabled me to make more of the subject than ,i could otherwise have done4 ,illustration3 #f ,illustration3 #g ,several correspondents have informed me that they possess1 or had possessed1 specimens of the old ,chinese books4 ,an ,american gentleman writes to me as follows3--,i have in my possession a book made of tissue paper1 printed in black 7with a ,chinese inscription on the front page71 containing over three hundred designs1 which belongs to the box of 'tangrams1' which ,i also own4 ,the blocks are seven in number1 made of mother-of-pearl1 highly polished and finely engraved on either side4 ,these are contained in a rosewood box #ba_/h in4 square4 ,my great uncle1 ----1 was one of the first missionaries to visit ,china4 ,this box and book1 along with quite a collection of other relics1 were sent to my grandfather and descended to myself4 ,my correspondent kindly supplied me with rubbings of the ,tangrams1 from which it is clear that they are cut in the exact proportions that ,i have indicated4 ,i reproduce the ,chinese inscription 7#h7 for this reason4 ,the owner of the book informs me that he has submitted it to a number of ,chinamen in the ,united ,states and offered as much as a dollar for a translation4 ,but they all steadfastly refused to read the words1 offering the lame excuse that the inscription is ,japanese4 ,natives of ,japan1 however1 insist that it is ,chinese4 ,is there something occult and esoteric about ,tangrams1 that it is so difficult to lift the veil8 ,perhaps this page will come under the eye of some reader acquainted with the ,chinese language1 who will supply the required translation1 which may1 or may not1 throw a little light on this curious question4 ,illustration3 #h ,by using several sets of ,tangrams at the same time we may construct more ambitious pictures4 ,i was advised by a friend not to send my picture1 ,a ,game of ,billiards 7#i71 to the ,academy4 ,he assured me that it would not be accepted because the judges are so hide-bound by convention4 ,perhaps he was right1 and it will be more appreciated by ,post-impressionists and ,cubists4 ,the players are considering a very delicate stroke at the top of the table4 ,of course1 the two men1 the table1 and the clock are formed from four sets of ,tangrams4 ,my second picture is named ,the ,orchestra 7#aj71 and it was designed for the decoration of a large hall of music4 ,here we have the conductor1 the pianist1 the fat little cornet-player1 the left-handed player of the double-bass1 whose attitude is life-like1 though he does stand at an unusual distance from his instrument1 and the drummer-boy1 with his imposing music-stand4 ,the dog at the back of the pianoforte is not howling3 he is an appreciative listener4 ,illustration3 #i ,illustration3 #aj ,one remarkable thing about these ,tangram pictures is that they suggest to the imagination such a lot that is not really there4 ,who1 for example1 can look for a few minutes at ,lady ,belinda 7#aa7 and the ,dutch girl 7#ab7 without soon feeling the haughty expression in the one case and the arch look in the other8 ,then look again at the stork 7#ac71 and see how it is suggested to the mind that the leg is actually much more slender than any one of the pieces employed4 ,it is really an optical illusion4 ,again1 notice in the case of the yacht 7#ad7 how1 by leaving that little angular point at the top1 a complete mast is suggested4 ,if you place your ,tangrams together on white paper so that they do not quite touch one another1 in some cases the effect is improved by the white lines2 in other cases it is almost destroyed4 ,illustration3 #aa ,illustration3 #ab ,finally1 ,i give an example from the many curious paradoxes that one happens upon in manipulating ,tangrams4 ,i show designs of two dignified individuals 7#ae and #af7 who appear to be exactly alike1 except for the fact that one has a foot and the other has not4 ,now1 both of these figures are made from the same seven ,tangrams4 ,where does the second man get his foot from8 ,illustration3 #ac ,illustration3 #ad ,illustration3 #ae ,illustration3 #af ,,patchwork ,,puzzles4 ,of shreds and patches4--,hamlet1 iii4 #d4 #agj4--,,the ,,cushion ,,covers4 ,illustration ,the above represents a square of brocade4 ,a lady wishes to cut it in four pieces so that two pieces will form one perfectly square cushion top1 and the remaining two pieces another square cushion top4 ,how is she to do it8 ,of course1 she can only cut along the lines that divide the twenty-five squares1 and the pattern must match properly without any irregularity whatever in the design of the material4 ,there is only one way of doing it4 ,can you find it8 #aga4--,,the ,,banner ,,puzzle4 ,illustration ,a ,lady had a square piece of bunting with two lions on it1 of which the illustration is an exactly reproduced reduction4 ,she wished to cut the stuff into pieces that would fit together and form two square banners with a lion on each banner4 ,she discovered that this could be done in as few as four pieces4 ,how did she manage it8 ,of course1 to cut the ,british ,lion would be an unpardonable offence1 so you must be careful that no cut passes through any portion of either of them4 ,ladies are informed that no allowance whatever has to be made for turnings1 and no part of the material may be wasted4 ,it is quite a simple little dissection puzzle if rightly attacked4 ,remember that the banners have to be perfect squares1 though they need not be both of the same size4 #agb4--,,mrs4 ,,smiley',s ,,christmas ,,present4 ,mrs4 ,smiley's expression of pleasure was sincere when her six granddaughters sent to her1 as a ,christmas present1 a very pretty patchwork quilt1 which they had made with their own hands4 ,it was constructed of square pieces of silk material1 all of one size1 and as they made a large quilt with fourteen of these little squares on each side1 it is obvious that just #aif pieces had been stitched into it4 ,now1 the six granddaughters each contributed a part of the work in the form of a perfect square 7all six portions being different in size71 but in order to join them up to form the square quilt it was necessary that the work of one girl should be unpicked into three separate pieces4 ,can you show how the joins might have been made8 ,of course1 no portion can be turned over4 ,illustration #agc4--,,mrs4 ,,perkins',s ,,quilt4 ,illustration ,it will be seen that in this case the square patchwork quilt is built up of #afi pieces4 ,the puzzle is to find the smallest possible number of square portions of which the quilt could be composed and show how they might be joined together4 ,or1 to put it the reverse way1 divide the quilt into as few square portions as possible by merely cutting the stitches4 #agd4--,,the ,,squares ,,of ,,brocade4 ,illustration ,i happened to be paying a call at the house of a lady1 when ,i took up from a table two lovely squares of brocade4 ,they were beautiful specimens of ,eastern workmanship--both of the same design1 a delicate chequered pattern4 ,are they not exquisite8 said my friend4 ,they were brought to me by a cousin who has just returned from ,india4 ,now1 ,i want you to give me a little assistance4 ,you see1 ,i have decided to join them together so as to make one large square cushion-cover4 ,how should ,i do this so as to mutilate the material as little as possible8 ,of course ,i propose to make my cuts only along the lines that divide the little chequers4 ,illustration ,i cut the two squares in the manner desired into four pieces that would fit together and form another larger square1 taking care that the pattern should match properly1 and when ,i had finished ,i noticed that two of the pieces were of exactly the same area2 that is1 each of the two contained the same number of chequers4 ,can you show how the cuts were made in accordance with these conditions8 #age--;,,another ,,patchwork ,,puzzle4 ,illustration ,a lady was presented1 by two of her girl friends1 with the pretty pieces of silk patchwork shown in our illustration4 ,it will be seen that both pieces are made up of squares all of the same size--one #ab #ab and the other #e #e4 ,she proposes to join them together and make one square patchwork quilt1 #ac #ac1 but1 of course1 she will not cut any of the material--merely cut the stitches where necessary and join together again4 ,what perplexes her is this4 ,a friend assures her that there need be no more than four pieces in all to join up for the new quilt4 ,could you show her how this little needlework puzzle is to be solved in so few pieces8 #agf4--,,linoleum ,,cutting4 ,illustration ,the diagram herewith represents two separate pieces of linoleum4 ,the chequered pattern is not repeated at the back1 so that the pieces cannot be turned over4 ,the puzzle is to cut the two squares into four pieces so that they shall fit together and form one perfect square #aj #aj1 so that the pattern shall properly match1 and so that the larger piece shall have as small a portion as possible cut from it4 #agg4--;,,another ,,linoleum ,,puzzle4 ,illustration ,can you cut this piece of linoleum into four pieces that will fit together and form a perfect square8 ,of course the cuts may only be made along the lines4 ,,various ,,geometrical ,,puzzles4 ,so various are the tastes of men4 ,,mark ,,akenside4 #agh4--,,the ,,cardboard ,,box4 ,this puzzle is not difficult1 but it will be found entertaining to discover the simple rule for its solution4 ,i have a rectangular cardboard box4 ,the top has an area of #abj square inches1 the side #if square inches1 and the end #hj square inches4 ,what are the exact dimensions of the box8 #agi4--,,stealing ,,the ,,bell-,,ropes4 ,two men broke into a church tower one night to steal the bell-ropes4 ,the two ropes passed through holes in the wooden ceiling high above them1 and they lost no time in climbing to the top4 ,then one man drew his knife and cut the rope above his head1 in consequence of which he fell to the floor and was badly injured4 ,his fellow-thief called out that it served him right for being such a fool4 ,he said that he should have done as he was doing1 upon which he cut the rope below the place at which he held on4 ,then1 to his dismay1 he found that he was in no better plight1 for1 after hanging on as long as his strength lasted1 he was compelled to let go and fall beside his comrade4 ,here they were both found the next morning with their limbs broken4 ,how far did they fall8 ,one of the ropes when they found it was just touching the floor1 and when you pulled the end to the wall1 keeping the rope taut1 it touched a point just three inches above the floor1 and the wall was four feet from the rope when it hung at rest4 ,how long was the rope from floor to ceiling8 #ahj4--,,the ,,four ,,sons4 ,readers will recognize the diagram as a familiar friend of their youth4 ,a man possessed a square-shaped estate4 ,he bequeathed to his widow the quarter of it that is shaded off4 ,the remainder was to be divided equitably amongst his four sons1 so that each should receive land of exactly the same area and exactly similar in shape4 ,we are shown how this was done4 ,but the remainder of the story is not so generally known4 ,in the centre of the estate was a well1 indicated by the dark spot1 and ,benjamin1 ,charles1 and ,david complained that the division was not equitable1 since ,alfred had access to this well1 while they could not reach it without trespassing on somebody else's land4 ,the puzzle is to show how the estate is to be apportioned so that each son shall have land of the same shape and area1 and each have access to the well without going off his own land4 ,illustration #aha4--,,the ,,three ,,railway ,,stations4 ,as ,i sat in a railway carriage ,i noticed at the other end of the compartment a worthy squire1 whom ,i knew by sight1 engaged in conversation with another passenger1 who was evidently a friend of his4 ,how far have you to drive to your place from the railway station8 asked the stranger4 ,well1 replied the squire1 if ,i get out at ,appleford1 it is just the same distance as if ,i go to ,bridgefield1 another fifteen miles farther on2 and if ,i changed at ,appleford and went thirteen miles from there to ,carterton1 it would still be the same distance4 ,you see1 ,i am equidistant from the three stations1 so ,i get a good choice of trains4 ,now ,i happened to know that ,bridgefield is just fourteen miles from ,carterton1 so ,i amused myself in working out the exact distance that the squire had to drive home whichever station he got out at4 ,what was the distance8 #ahb4--,,the ,,garden ,,puzzle4 ,professor ,rackbrain tells me that he was recently smoking a friendly pipe under a tree in the garden of a country acquaintance4 ,the garden was enclosed by four straight walls1 and his friend informed him that he had measured these and found the lengths to be #hj1 #de1 #ajj1 and #fc yards respectively4 ,then1 said the professor1 we can calculate the exact area of the garden4 ,impossible1 his host replied1 because you can get an infinite number of different shapes with those four sides4 ,but you forget1 ,rackbrane said1 with a twinkle in his eye1 that you told me once you had planted this tree equidistant from all the four corners of the garden4 ,can you work out the garden's area8 #ahc4--;,,drawing ,a ,,spiral4 ,if you hold the page horizontally and give it a quick rotary motion while looking at the centre of the spiral1 it will appear to revolve4 ,perhaps a good many readers are acquainted with this little optical illusion4 ,but the puzzle is to show how ,i was able to draw this spiral with so much exactitude without using anything but a pair of compasses and the sheet of paper on which the diagram was made4 ,how would you proceed in such circumstances8 ,illustration #ahd4--;,,how ,,to ,,draw ,,an ,,oval4 ,can you draw a perfect oval on a sheet of paper with one sweep of the compasses8 ,it is one of the easiest things in the world when you know how4 #ahe4--,,st4 ,,george',s ,,banner4 ,at a celebration of the national festival of ,st4 ,george's ,day ,i was contemplating the familiar banner of the patron saint of our country4 ,we all know the red cross on a white ground1 shown in our illustration4 ,this is the banner of ,st4 ,george4 ,the banner of ,st4 ,andrew 7,scotland7 is a white ,st4 ,andrew's ,cross on a blue ground4 ,that of ,st4 ,patrick 7,ireland7 is a similar cross in red on a white ground4 ,these three are united in one to form our ,union ,jack4 ,now on looking at ,st4 ,george's banner it occurred to me that the following question would make a simple but pretty little puzzle4 ,supposing the flag measures four feet by three feet1 how wide must the arm of the cross be if it is required that there shall be used just the same quantity of red and of white bunting8 ,illustration #ahf4--,,the ,,clothes ,,line ,,puzzle4 ,a boy tied a clothes line from the top of each of two poles to the base of the other4 ,he then proposed to his father the following question4 ,as one pole was exactly seven feet above the ground and the other exactly five feet1 what was the height from the ground where the two cords crossed one another8 #ahg4--,,the ,,milkmaid ,,puzzle4 ,illustration ,here is a little pastoral puzzle that the reader may1 at first sight1 be led into supposing is very profound1 involving deep calculations4 ,he may even say that it is quite impossible to give any answer unless we are told something definite as to the distances4 ,and yet it is really quite childlike and bland4 ,in the corner of a field is seen a milkmaid milking a cow1 and on the other side of the field is the dairy where the extract has to be deposited4 ,but it has been noticed that the young woman always goes down to the river with her pail before returning to the dairy4 ,here the suspicious reader will perhaps ask why she pays these visits to the river4 ,i can only reply that it is no business of ours4 ,the alleged milk is entirely for local consumption4 ,where are you going to1 my pretty maid8 ,down to the river1 sir1 she said4 ,i'll not choose your dairy1 my pretty maid4 ,nobody axed you1 sir1 she said4 ,if one had any curiosity in the matter1 such an independent spirit would entirely disarm one4 ,so we will pass from the point of commercial morality to the subject of the puzzle4 ,draw a line from the milking-stool down to the river and thence to the door of the dairy1 which shall indicate the shortest possible route for the milkmaid4 ,that is all4 ,it is quite easy to indicate the exact spot on the bank of the river to which she should direct her steps if she wants as short a walk as possible4 ,can you find that spot8 #ahh4--,,the ,,ball ,,problem4 ,illustration ,a stonemason was engaged the other day in cutting out a round ball for the purpose of some architectural decoration1 when a smart schoolboy came upon the scene4 ,look here1 said the mason1 you seem to be a sharp youngster1 can you tell me this8 ,if ,i placed this ball on the level ground1 how many other balls of the same size could ,i lay around it 7also on the ground7 so that every ball should touch this one8 ,the boy at once gave the correct answer1 and then put this little question to the mason3-- ,if the surface of that ball contained just as many square feet as its volume contained cubic feet1 what would be the length of its diameter8 ,the stonemason could not give an answer4 ,could you have replied correctly to the mason's and the boy's questions8 #ahi4--,,the ,,yorkshire ,,estates4 ,illustration ,i was on a visit to one of the large towns of ,yorkshire4 ,while walking to the railway station on the day of my departure a man thrust a hand-bill upon me1 and ,i took this into the railway carriage and read it at my leisure4 ,it informed me that three ,yorkshire neighbouring estates were to be offered for sale4 ,each estate was square in shape1 and they joined one another at their corners1 just as shown in the diagram4 ,estate ,a contains exactly #cgj acres1 ,b contains #aaf acres1 and ,c #gd acres4 ,now1 the little triangular bit of land enclosed by the three square estates was not offered for sale1 and1 for no reason in particular1 ,i became curious as to the area of that piece4 ,how many acres did it contain8 #aij4--;,,farmer ,,wurzel',s ,,estate4 ,illustration ,i will now present another land problem4 ,the demonstration of the answer that ,i shall give will1 ,i think1 be found both interesting and easy of comprehension4 ,farmer ,wurzel owned the three square fields shown in the annexed plan1 containing respectively #ah1 #bj1 and #bf acres4 ,in order to get a ring-fence round his property he bought the four intervening triangular fields4 ,the puzzle is to discover what was then the whole area of his estate4 #aia4--,,the ,,crescent ,,puzzle4 ,illustration ,here is an easy geometrical puzzle4 ,the crescent is formed by two circles1 and ,c is the centre of the larger circle4 ,the width of the crescent between ,b and ,d is #i inches1 and between ,e and ,f #e inches4 ,what are the diameters of the two circles8 #aib4--,,the ,,puzzle ,,wall4 ,illustration ,there was a small lake1 around which four poor men built their cottages4 ,four rich men afterwards built their mansions1 as shown in the illustration1 and they wished to have the lake to themselves1 so they instructed a builder to put up the shortest possible wall that would exclude the cottagers1 but give themselves free access to the lake4 ,how was the wall to be built8 #aic4--,,the ,,sheepfold4 ,it is a curious fact that the answers always given to some of the best-known puzzles that appear in every little book of fireside recreations that has been published for the last fifty or a hundred years are either quite unsatisfactory or clearly wrong4 ,yet nobody ever seems to detect their faults4 ,here is an example3--,a farmer had a pen made of fifty hurdles1 capable of holding a hundred sheep only4 ,supposing he wanted to make it sufficiently large to hold double that number1 how many additional hurdles must he have8 #aid4--,,the ,,garden ,,walls4 ,illustration ,a speculative country builder has a circular field1 on which he has erected four cottages1 as shown in the illustration4 ,the field is surrounded by a brick wall1 and the owner undertook to put up three other brick walls1 so that the neighbours should not be overlooked by each other1 but the four tenants insist that there shall be no favouritism1 and that each shall have exactly the same length of wall space for his wall fruit trees4 ,the puzzle is to show how the three walls may be built so that each tenant shall have the same area of ground1 and precisely the same length of wall4 ,of course1 each garden must be entirely enclosed by its walls1 and it must be possible to prove that each garden has exactly the same length of wall4 ,if the puzzle is properly solved no figures are necessary4 #aie4--,,lady ,,belinda',s ,,garden4 ,lady ,belinda is an enthusiastic gardener4 ,in the illustration she is depicted in the act of worrying out a pleasant little problem which ,i will relate4 ,one of her gardens is oblong in shape1 enclosed by a high holly hedge1 and she is turning it into a rosary for the cultivation of some of her choicest roses4 ,she wants to devote exactly half of the area of the garden to the flowers1 in one large bed1 and the other half to be a path going all round it of equal breadth throughout4 ,such a garden is shown in the diagram at the foot of the picture4 ,how is she to mark out the garden under these simple conditions8 ,she has only a tape1 the length of the garden1 to do it with1 and1 as the holly hedge is so thick and dense1 she must make all her measurements inside4 ,lady ,belinda did not know the exact dimensions of the garden1 and1 as it was not necessary for her to know1 ,i also give no dimensions4 ,it is quite a simple task no matter what the size or proportions of the garden may be4 ,yet how many lady gardeners would know just how to proceed8 ,the tape may be quite plain--that is1 it need not be a graduated measure4 ,illustration #aif4--,,the ,,tethered ,,goat4 ,illustration ,here is a little problem that everybody should know how to solve4 ,the goat is placed in a half-acre meadow1 that is in shape an equilateral triangle4 ,it is tethered to a post at one corner of the field4 ,what should be the length of the tether 7to the nearest inch7 in order that the goat shall be able to eat just half the grass in the field8 ,it is assumed that the goat can feed to the end of the tether4 #aig4--,,the ,,compasses ,,puzzle4 ,it is curious how an added condition or restriction will sometimes convert an absurdly easy puzzle into an interesting and perhaps difficult one4 ,i remember buying in the street many years ago a little mechanical puzzle that had a tremendous sale at the time4 ,it consisted of a medal with holes in it1 and the puzzle was to work a ring with a gap in it from hole to hole until it was finally detached4 ,as ,i was walking along the street ,i very soon acquired the trick of taking off the ring with one hand while holding the puzzle in my pocket4 ,a friend to whom ,i showed the little feat set about accomplishing it himself1 and when ,i met him some days afterwards he exhibited his proficiency in the art4 ,but he was a little taken aback when ,i then took the puzzle from him and1 while simply holding the medal between the finger and thumb of one hand1 by a series of little shakes and jerks caused the ring1 without my even touching it1 to fall off upon the floor4 ,the following little poser will probably prove a rather tough nut for a great many readers1 simply on account of the restricted conditions3-- ,show how to find exactly the middle of any straight line by means of the compasses only4 ,you are not allowed to use any ruler1 pencil1 or other article--only the compasses2 and no trick or dodge1 such as folding the paper1 will be permitted4 ,you must simply use the compasses in the ordinary legitimate way4 #aih4--,,the ,,eight ,,sticks4 ,i have eight sticks1 four of them being exactly half the length of the others4 ,i lay every one of these on the table1 so that they enclose three squares1 all of the same size4 ,how do ,i do it8 ,there must be no loose ends hanging over4 #aii4--,,papa',s ,,puzzle4 ,here is a puzzle by ,pappus1 who lived at ,alexandria about the end of the third century4 ,it is the fifth proposition in the eighth book of his ,mathematical ,collections4 ,i give it in the form that ,i presented it some years ago under the title ,papa's ,puzzle1 just to see how many readers would discover that it was by ,pappus himself4 ,the little maid's papa has taken two different-sized rectangular pieces of cardboard1 and has clipped off a triangular piece from one of them1 so that when it is suspended by a thread from the point ,a it hangs with the long side perfectly horizontal1 as shown in the illustration4 ,he has perplexed the child by asking her to find the point ,a on the other card1 so as to produce a similar result when cut and suspended by a thread4 ,of course1 the point must not be found by trial clippings4 ,a curious and pretty point is involved in this setting of the puzzle4 ,can the reader discover it8 ,illustration #bjj4--;,a ,,kite-,,flying ,,puzzle4 ,while accompanying my friend ,professor ,highflite during a scientific kite-flying competition on the ,south ,downs of ,sussex ,i was led into a little calculation that ought to interest my readers4 ,the ,professor was paying out the wire to which his kite was attached from a winch on which it had been rolled into a perfectly spherical form4 ,this ball of wire was just two feet in diameter1 and the wire had a diameter of one-hundredth of an inch4 ,what was the length of the wire8 ,now1 a simple little question like this that everybody can perfectly understand will puzzle many people to answer in any way4 ,let us see whether1 without going into any profound mathematical calculations1 we can get the answer roughly--say1 within a mile of what is correct6 ,we will assume that when the wire is all wound up the ball is perfectly solid throughout1 and that no allowance has to be made for the axle that passes through it4 ,with that simplification1 ,i wonder how many readers can state within even a mile of the correct answer the length of that wire4 #bja4--;,,how ,,to ,,make ,,cisterns4 ,illustration ,our friend in the illustration has a large sheet of zinc1 measuring 7before cutting7 eight feet by three feet1 and he has cut out square pieces 7all of the same size7 from the four corners and now proposes to fold up the sides1 solder the edges1 and make a cistern4 ,but the point that puzzles him is this3 ,has he cut out those square pieces of the correct size in order that the cistern may hold the greatest possible quantity of water8 ,you see1 if you cut them very small you get a very shallow cistern2 if you cut them large you get a tall and slender one4 ,it is all a question of finding a way of cutting put these four square pieces exactly the right size4 ,how are we to avoid making them too small or too large8 #bjb4--,,the ,,cone ,,puzzle4 ,illustration ,i have a wooden cone1 as shown in ,fig4 #a4 ,how am ,i to cut out of it the greatest possible cylinder8 ,it will be seen that ,i can cut out one that is long and slender1 like ,fig4 #b1 or short and thick1 like ,fig4 #c4 ,but neither is the largest possible4 ,a child could tell you where to cut1 if he knew the rule4 ,can you find this simple rule8 #bjc4--;,,concerning ,,wheels4 ,illustration ,there are some curious facts concerning the movements of wheels that are apt to perplex the novice4 ,for example3 when a railway train is travelling from ,london to ,crewe certain parts of the train at any given moment are actually moving from ,crewe towards ,london4 ,can you indicate those parts8 ,it seems absurd that parts of the same train can at any time travel in opposite directions1 but such is the case4 ,in the accompanying illustration we have two wheels4 ,the lower one is supposed to be fixed and the upper one running round it in the direction of the arrows4 ,now1 how many times does the upper wheel turn on its own axis in making a complete revolution of the other wheel8 ,do not be in a hurry with your answer1 or you are almost certain to be wrong4 ,experiment with two pennies on the table and the correct answer will surprise you1 when you succeed in seeing it4 #bjd4--;,a ,,new ,,match ,,puzzle4 ,illustration ,in the illustration eighteen matches are shown arranged so that they enclose two spaces1 one just twice as large as the other4 ,can you rearrange them 7#a7 so as to enclose two four-sided spaces1 one exactly three times as large as the other1 and 7#b7 so as to enclose two five-sided spaces1 one exactly three times as large as the other8 ,all the eighteen matches must be fairly used in each case2 the two spaces must be quite detached1 and there must be no loose ends or duplicated matches4 #bje4--,,the ,,six ,,sheep-,,pens4 ,illustration ,here is a new little puzzle with matches4 ,it will be seen in the illustration that thirteen matches1 representing a farmer's hurdles1 have been so placed that they enclose six sheep-pens all of the same size4 ,now1 one of these hurdles was stolen1 and the farmer wanted still to enclose six pens of equal size with the remaining twelve4 ,how was he to do it8 ,all the twelve matches must be fairly used1 and there must be no duplicated matches or loose ends4 ,,points ,,and ,,lines ,,problems4 ,line upon line1 line upon line2 here a little and there a little4--,isa4 xxviii4 #aj4 ,what are known as ,points and ,lines puzzles are found very interesting by many people4 ,the most familiar example1 here given1 to plant nine trees so that they shall form ten straight rows with three trees in every row1 is attributed to ,sir ,isaac ,newton1 but the earliest collection of such puzzles is1 ,i believe1 in a rare little book that ,i possess--published in #ahba--,rational ,amusement for ,winter ,evenings1 by ,john ,jackson4 ,the author gives ten examples of ,trees planted in ,rows4 ,these tree-planting puzzles have always been a matter of great perplexity4 ,they are real puzzles1 in the truest sense of the word1 because nobody has yet succeeded in finding a direct and certain way of solving them4 ,they demand the exercise of sagacity1 ingenuity1 and patience1 and what we call luck is also sometimes of service4 ,perhaps some day a genius will discover the key to the whole mystery4 ,remember that the trees must be regarded as mere points1 for if we were allowed to make our trees big enough we might easily fudge our diagrams and get in a few extra straight rows that were more apparent than real4 ,illustration #bjf4--,,the ,,king ,,and ,,the ,,castles4 ,there was once1 in ancient times1 a powerful king1 who had eccentric ideas on the subject of military architecture4 ,he held that there was great strength and economy in symmetrical forms1 and always cited the example of the bees1 who construct their combs in perfect hexagonal cells1 to prove that he had nature to support him4 ,he resolved to build ten new castles in his country all to be connected by fortified walls1 which should form five lines with four castles in every line4 ,the royal architect presented his preliminary plan in the form ,i have shown4 ,but the monarch pointed out that every castle could be approached from the outside1 and commanded that the plan should be so modified that as many castles as possible should be free from attack from the outside1 and could only be reached by crossing the fortified walls4 ,the architect replied that he thought it impossible so to arrange them that even one castle1 which the king proposed to use as a royal residence1 could be so protected1 but his majesty soon enlightened him by pointing out how it might be done4 ,how would you have built the ten castles and fortifications so as best to fulfil the king's requirements8 ,remember that they must form five straight lines with four castles in every line4 ,illustration #bjg4--;,,cherries ,,and ,,plums4 ,illustration ,the illustration is a plan of a cottage as it stands surrounded by an orchard of fifty-five trees4 ,ten of these trees are cherries1 ten are plums1 and the remainder apples4 ,the cherries are so planted as to form five straight lines1 with four cherry trees in every line4 ,the plum trees are also planted so as to form five straight lines with four plum trees in every line4 ,the puzzle is to show which are the ten cherry trees and which are the ten plums4 ,in order that the cherries and plums should have the most favourable aspect1 as few as possible 7under the conditions7 are planted on the north and east sides of the orchard4 ,of course in picking out a group of ten trees 7cherry or plum1 as the case may be7 you ignore all intervening trees4 ,that is to say1 four trees may be in a straight line irrespective of other trees 7or the house7 being in between4 ,after the last puzzle this will be quite easy4 #bjh4--;,a ,,plantation ,,puzzle4 ,illustration ,a man had a square plantation of forty-nine trees1 but1 as will be seen by the omissions in the illustration1 four trees were blown down and removed4 ,he now wants to cut down all the remainder except ten trees1 which are to be so left that they shall form five straight rows with four trees in every row4 ,which are the ten trees that he must leave8 #bji4--,,the ,,twenty-,,one ,,trees4 ,a gentleman wished to plant twenty-one trees in his park so that they should form twelve straight rows with five trees in every row4 ,could you have supplied him with a pretty symmetrical arrangement that would satisfy these conditions8 #baj4--,,the ,,ten ,,coins4 ,place ten pennies on a large sheet of paper or cardboard1 as shown in the diagram1 five on each edge4 ,now remove four of the coins1 without disturbing the others1 and replace them on the paper so that the ten shall form five straight lines with four coins in every line4 ,this in itself is not difficult1 but you should try to discover in how many different ways the puzzle may be solved1 assuming that in every case the two rows at starting are exactly the same4 ,illustration #baa4--,,the ,,twelve ,,mince-,,pies4 ,it will be seen in our illustration how twelve mince-pies may be placed on the table so as to form six straight rows with four pies in every row4 ,the puzzle is to remove only four of them to new positions so that there shall be seven straight rows with four in every row4 ,which four would you remove1 and where would you replace them8 ,illustration #bab4--,,the ,,burmese ,,plantation4 ,illustration ,a short time ago ,i received an interesting communication from the ,british chaplain at ,meiktila1 ,upper ,burma1 in which my correspondent informed me that he had found some amusement on board ship on his way out in trying to solve this little poser4 ,if he has a plantation of forty-nine trees1 planted in the form of a square as shown in the accompanying illustration1 he wishes to know how he may cut down twenty-seven of the trees so that the twenty-two left standing shall form as many rows as possible with four trees in every row4 ,of course there may not be more than four trees in any row4 #bac4--,,turks ,,and ,,russians4 ,this puzzle is on the lines of the ,afridi problem published by me in ,tit-,bits some years ago4 ,on an open level tract of country a party of ,russian infantry1 no two of whom were stationed at the same spot1 were suddenly surprised by thirty-two ,turks1 who opened fire on the ,russians from all directions4 ,each of the ,turks simultaneously fired a bullet1 and each bullet passed immediately over the heads of three ,russian soldiers4 ,as each of these bullets when fired killed a different man1 the puzzle is to discover what is the smallest possible number of soldiers of which the ,russian party could have consisted and what were the casualties on each side4 ,,moving ,,counter ,,problems4 ,i cannot do't without counters4 ,winter's ,tale1 iv4 #c4 ,puzzles of this class1 except so far as they occur in connection with actual games1 such as chess1 seem to be a comparatively modern introduction4 ,mathematicians in recent times1 notably ,vandermonde and ,reiss1 have devoted some attention to them1 but they do not appear to have been considered by the old writers4 ,so far as games with counters are concerned1 perhaps the most ancient and widely known in old times is ,nine ,men's ,morris 7known also1 as ,i shall show1 under a great many other names71 unless the simpler game1 distinctly mentioned in the works of ,ovid 7,no4 #aaj1 ,ovid's ,game1 in ,the ,canterbury ,puzzles71 from which ,noughts and ,crosses seems to be derived1 is still more ancient4 ,in ,france the game is called ,marelle1 in ,poland ,siegen ,wulf ,myll 7,she-goat ,wolf ,mill1 or ,fight71 in ,germany and ,austria it is called ,muhle 7the ,mill71 in ,iceland it goes by the name of ,mylla1 while the ,bogas 7or native bargees7 of ,south ,america are said to play it1 and on the ,amazon it is called ,trique1 and held to be of ,indian origin4 ,in our own country it has different names in different districts1 such as ,meg ,merrylegs1 ,peg ,meryll1 ,nine ,peg o',merryal1 ,nine-,pin ,miracle1 ,merry ,peg1 and ,merry ,hole4 ,shakespeare refers to it in ,midsummer ,night's ,dream 7,act ii41 scene #a73-- ,the nine-men's morris is filled up with mud2 ,and the quaint mazes in the wanton green1 ,for lack of tread1 are undistinguishable4 ,it was played by the shepherds with stones in holes cut in the turf4 ,john ,clare1 the peasant poet of ,northamptonshire1 in ,the ,shepherd ,boy 7#ahce7 says3--,oft we track his haunts 4444 ,by nine-peg-morris nicked upon the green4 ,it is also mentioned by ,drayton in his ,polyolbion4 ,it was found on an old ,roman tile discovered during the excavations at ,silchester1 and cut upon the steps of the ,acropolis at ,athens4 ,when visiting the ,christiania ,museum a few years ago ,i was shown the great ,viking ship that was discovered at ,gokstad in #ahhj4 ,on the oak planks forming the deck of the vessel were found boles and lines marking out the game1 the holes being made to receive pegs4 ,while inspecting the ancient oak furniture in the ,rijks ,museum at ,amsterdam ,i became interested in an old catechumen's settle1 and was surprised to find the game diagram cut in the centre of the seat--quite conveniently for surreptitious play4 ,it has been discovered cut in the choir stalls of several of our ,english cathedrals4 ,in the early eighties it was found scratched upon a stone built into a wall 7probably about the date #abjj71 during the restoration of ,hargrave church in ,northamptonshire4 ,this stone is now in the ,northampton ,museum4 ,a similar stone has since been found at ,sempringham1 ,lincolnshire4 ,it is to be seen on an ancient tombstone in the ,isle of ,man1 and painted on old ,dutch tiles4 ,and in #aija a stone was dug out of a gravel pit near ,oswestry bearing an undoubted diagram of the game4 ,the game has been played with different rules at different periods and places4 ,i give a copy of the board4 ,sometimes the diagonal lines are omitted1 but this evidently was not intended to affect the play3 it simply meant that the angles alone were thought sufficient to indicate the points4 ,this is how ,strutt1 in ,sports and ,pastimes1 describes the game1 and it agrees with the way ,i played it as a boy3--,two persons1 having each of them nine pieces1 or men1 lay them down alternately1 one by one1 upon the spots2 and the business of either party is to prevent his antagonist from placing three of his pieces so as to form a row of three1 without the intervention of an opponent piece4 ,if a row be formed1 he that made it is at liberty to take up one of his competitor's pieces from any part he thinks most to his advantage2 excepting he has made a row1 which must not be touched if he have another piece upon the board that is not a component part of that row4 ,when all the pieces are laid down1 they are played backwards and forwards1 in any direction that the lines run1 but only can move from one spot to another 7next to it7 at one time4 ,he that takes off all his antagonist's pieces is the conqueror4 ,illustration #bad4--,,the ,,six ,,frogs4 ,illustration ,the six educated frogs in the illustration are trained to reverse their order1 so that their numbers shall read #f1 #e1 #d1 #c1 #b1 #a1 with the blank square in its present position4 ,they can jump to the next square 7if vacant7 or leap over one frog to the next square beyond 7if vacant71 just as we move in the game of draughts1 and can go backwards or forwards at pleasure4 ,can you show how they perform their feat in the fewest possible moves8 ,it is quite easy1 so when you have done it add a seventh frog to the right and try again4 ,then add more frogs until you are able to give the shortest solution for any number4 ,for it can always be done1 with that single vacant square1 no matter how many frogs there are4 #bae4--,,the ,,grasshopper ,,puzzle4 ,it has been suggested that this puzzle was a great favourite among the young apprentices of the ,city of ,london in the sixteenth and seventeenth centuries4 ,readers will have noticed the curious brass grasshopper on the ,royal ,exchange4 ,this long-lived creature escaped the fires of #afff and #ahch4 ,the grasshopper1 after his kind1 was the crest of ,sir ,thomas ,gresham1 merchant grocer1 who died in #aegi1 and from this cause it has been used as a sign by grocers in general4 ,unfortunately for the legend as to its origin1 the puzzle was only produced by myself so late as the year #aijj4 ,on twelve of the thirteen black discs are placed numbered counters or grasshoppers4 ,the puzzle is to reverse their order1 so that they shall read1 #a1 #b1 #c1 #d1 etc41 in the opposite direction1 with the vacant disc left in the same position as at present4 ,move one at a time in any order1 either to the adjoining vacant disc or by jumping over one grasshopper1 like the moves in draughts4 ,the moves or leaps may be made in either direction that is at any time possible4 ,what are the fewest possible moves in which it can be done8 ,illustration #baf4--,,the ,,educated ,,frogs4 ,illustration ,our six educated frogs have learnt a new and pretty feat4 ,when placed on glass tumblers1 as shown in the illustration1 they change sides so that the three black ones are to the left and the white frogs to the right1 with the unoccupied tumbler at the opposite end--,no4 #g4 ,they can jump to the next tumbler 7if unoccupied71 or over one1 or two1 frogs to an unoccupied tumbler4 ,the jumps can be made in either direction1 and a frog may jump over his own or the opposite colour1 or both colours4 ,four successive specimen jumps will make everything quite plain3 #d to #a1 #e to #d1 #c to #e1 #f to #c4 ,can you show how they do it in ten jumps8 #bag4--,,the ,,twickenham ,,puzzle4 ,illustration3 7 ,i 7 77,n77 7 ,m 7 77,a77 7 ,h 7 77,t77 7 ,e 7 77,w77 7 ,c 7 77,k77 7 7 ,in the illustration we have eleven discs in a circle4 ,on five of the discs we place white counters with black letters--as shown--and on five other discs the black counters with white letters4 ,the bottom disc is left vacant4 ,starting thus1 it is required to get the counters into order so that they spell the word ,twickenham in a clockwise direction1 leaving the vacant disc in the original position4 ,the black counters move in the direction that a clock-hand revolves1 and the white counters go the opposite way4 ,a counter may jump over one of the opposite colour if the vacant disc is next beyond4 ,thus1 if your first move is with ,k1 then ,c can jump over ,k4 ,if then ,k moves towards ,e1 you may next jump ,w over ,c1 and so on4 ,the puzzle may be solved in twenty-six moves4 ,remember a counter cannot jump over one of its own colour4 #bah4--,,the ,,victoria ,,cross ,,puzzle4 ,illustration3 --------------------- 444 ,a 444_/ 7,i7 4444444 7,v7 _/ 444444 ------ 44 ,r 4 4 ,i 44 444444 444444 _/ 4444444 7,o7 44 ,t 44 7,c7 _/444444444 --------------------- ,the puzzle-maker is peculiarly a snapper-up of unconsidered trifles1 and his productions are often built up with the slenderest materials4 ,trivialities that might entirely escape the observation of others1 or1 if they were observed1 would be regarded as of no possible moment1 often supply the man who is in quest of posers with a pretty theme or an idea that he thinks possesses some basal value4 ,when seated opposite to a lady in a railway carriage at the time of ,queen ,victoria's ,diamond ,jubilee1 my attention was attracted to a brooch that she was wearing4 ,it was in the form of a ,maltese or ,victoria ,cross1 and bore the letters of the word ,,victoria4 ,the number and arrangement of the letters immediately gave me the suggestion for the puzzle which ,i now present4 ,the diagram1 it will be seen1 is composed of nine divisions4 ,the puzzle is to place eight counters1 bearing the letters of the word ,,victoria1 exactly in the manner shown1 and then slide one letter at a time from black to white and white to black alternately1 until the word reads round in the same direction1 only with the initial letter ,v on one of the black arms of the cross4 ,at no time may two letters be in the same division4 ,it is required to find the shortest method4 ,leaping moves are1 of course1 not permitted4 ,the first move must obviously be made with ,a1 ,i1 ,t1 or ,r4 ,supposing you move ,t to the centre1 the next counter played will be ,o or ,c1 since ,i or ,r cannot be moved4 ,there is something a little remarkable in the solution of this puzzle which ,i will explain4 #bai4--,,the ,,letter ,,block ,,puzzle4 ,illustration3 --------------- ,g ,e ,f --------------- ,h ,c ,b --------------- ,d ,a --------------- ,here is a little reminiscence of our old friend the ,fifteen ,block ,puzzle4 ,eight wooden blocks are lettered1 and are placed in a box1 as shown in the illustration4 ,it will be seen that you can only move one block at a time to the place vacant for the time being1 as no block may be lifted out of the box4 ,the puzzle is to shift them about until you get them in the order-- ,a ,b ,c ,d ,e ,f ,g ,h ,this you will find by no means difficult if you are allowed as many moves as you like4 ,but the puzzle is to do it in the fewest possible moves4 ,i will not say what this smallest number of moves is1 because the reader may like to discover it for himself4 ,in writing down your moves you will find it necessary to record no more than the letters in the order that they are shifted4 ,thus1 your first five moves might be ,c1 ,h1 ,g1 ,e1 ,f2 and this notation can have no possible ambiguity4 ,in practice you only need eight counters and a simple diagram on a sheet of paper4 #bbj4--;,a ,,lodging-,,house ,,difficulty4 ,illustration ,the ,dobsons secured apartments at ,slocomb-on-,sea4 ,there were six rooms on the same floor1 all communicating1 as shown in the diagram4 ,the rooms they took were numbers #d1 #e1 and #f1 all facing the sea4 ,but a little difficulty arose4 ,mr4 ,dobson insisted that the piano and the bookcase should change rooms4 ,this was wily1 for the ,dobsons were not musical1 but they wanted to prevent any one else playing the instrument4 ,now1 the rooms were very small and the pieces of furniture indicated were very big1 so that no two of these articles could be got into any room at the same time4 ,how was the exchange to be made with the least possible labour8 ,suppose1 for example1 you first move the wardrobe into ,no4 #b2 then you can move the bookcase to ,no4 #e and the piano to ,no4 #f1 and so on4 ,it is a fascinating puzzle1 but the landlady had reasons for not appreciating it4 ,try to solve her difficulty in the fewest possible removals with counters on a sheet of paper4 #bba4--,,the ,,eight ,,engines4 ,the diagram represents the engine-yard of a railway company under eccentric management4 ,the engines are allowed to be stationary only at the nine points indicated1 one of which is at present vacant4 ,it is required to move the engines1 one at a time1 from point to point1 in seventeen moves1 so that their numbers shall be in numerical order round the circle1 with the central point left vacant4 ,but one of the engines has had its fire drawn1 and therefore cannot move4 ,how is the thing to be done8 ,and which engine remains stationary throughout8 ,illustration #bbb4--;,a ,,railway ,,puzzle4 ,illustration ,make a diagram1 on a large sheet of paper1 like the illustration1 and have three counters marked ,a1 three marked ,b1 and three marked ,c4 ,it will be seen that at the intersection of lines there are nine stopping-places1 and a tenth stopping-place is attached to the outer circle like the tail of a ,q4 ,place the three counters or engines marked ,a1 the three marked ,b1 and the three marked ,c at the places indicated4 ,the puzzle is to move the engines1 one at a time1 along the lines1 from stopping-place to stopping-place1 until you succeed in getting an ,a1 a ,b1 and a ,c on each circle1 and also ,a1 ,b1 and ,c on each straight line4 ,you are required to do this in as few moves as possible4 ,how many moves do you need8 #bbc4--;,a ,,railway ,,muddle4 ,the plan represents a portion of the line of the ,london1 ,clodville1 and ,mudford ,railway ,company4 ,it is a single line with a loop4 ,there is only room for eight wagons1 or seven wagons and an engine1 between ,b and ,c on either the left line or the right line of the loop4 ,it happened that two goods trains 7each consisting of an engine and sixteen wagons7 got into the position shown in the illustration4 ,it looked like a hopeless deadlock1 and each engine-driver wanted the other to go back to the next station and take off nine wagons4 ,but an ingenious stoker undertook to pass the trains and send them on their respective journeys with their engines properly in front4 ,he also contrived to reverse the engines the fewest times possible4 ,could you have performed the feat8 ,and how many times would you require to reverse the engines8 ,a reversal means a change of direction1 backward or forward4 ,no rope-shunting1 fly-shunting1 or other trick is allowed4 ,all the work must be done legitimately by the two engines4 ,it is a simple but interesting puzzle if attempted with counters4 ,illustration #bbd4--,,the ,,motor-,,garage ,,puzzle4 ,illustration ,the difficulties of the proprietor of a motor garage are converted into a little pastime of a kind that has a peculiar fascination4 ,all you need is to make a simple plan or diagram on a sheet of paper or cardboard and number eight counters1 #a to #h4 ,then a whole family can enter into an amusing competition to find the best possible solution of the difficulty4 ,the illustration represents the plan of a motor garage1 with accommodation for twelve cars4 ,but the premises are so inconveniently restricted that the proprietor is often caused considerable perplexity4 ,suppose1 for example1 that the eight cars numbered #a to #h are in the positions shown1 how are they to be shifted in the quickest possible way so that #a1 #b1 #c1 and #d shall change places with #e1 #f1 #g1 and #h--that is1 with the numbers still running from left to right1 as at present1 but the top row exchanged with the bottom row8 ,what are the fewest possible moves8 ,one car moves at a time1 and any distance counts as one move4 ,to prevent misunderstanding1 the stopping-places are marked in squares1 and only one car can be in a square at the same time4 #bbe4--,,the ,,ten ,,prisoners4 ,if prisons had no other use1 they might still be preserved for the special benefit of puzzle-makers4 ,they appear to be an inexhaustible mine of perplexing ideas4 ,here is a little poser that will perhaps interest the reader for a short period4 ,we have in the illustration a prison of sixteen cells4 ,the locations of the ten prisoners will be seen4 ,the jailer has queer superstitions about odd and even numbers1 and he wants to rearrange the ten prisoners so that there shall be as many even rows of men1 vertically1 horizontally1 and diagonally1 as possible4 ,at present it will be seen1 as indicated by the arrows1 that there are only twelve such rows of #b and #d4 ,i will state at once that the greatest number of such rows that is possible is sixteen4 ,but the jailer only allows four men to be removed to other cells1 and informs me that1 as the man who is seated in the bottom right-hand corner is infirm1 he must not be moved4 ,now1 how are we to get those sixteen rows of even numbers under such conditions8 ,illustration #bbf4--,,round ,,the ,,coast4 ,illustration ,here is a puzzle that will1 ,i think1 be found as amusing as instructive4 ,we are given a ring of eight circles4 ,leaving circle #h blank1 we are required to write in the name of a seven-lettered port in the ,united ,kingdom in this manner4 ,touch a blank circle with your pencil1 then jump over two circles in either direction round the ring1 and write down the first letter4 ,then touch another vacant circle1 jump over two circles1 and write down your second letter4 ,proceed similarly with the other letters in their proper order until you have completed the word4 ,thus1 suppose we select ,glasgow1 and proceed as follows3 #f--a1 #g--b1 #h--c1 #g--d1 #h--e1 which means that we touch #f1 jump over #g and and write down ,g on #a2 then touch #g1 jump over #h and #a1 and write down l on #b2 and so on4 ,it will be found that after we have written down the first five letters--,glasg--as above1 we cannot go any further4 ,either there is something wrong with ,glasgow1 or we have not managed our jumps properly4 ,can you get to the bottom of the mystery8 #bbg4--;,,central ,,solitaire4 ,illustration ,this ancient puzzle was a great favourite with our grandmothers1 and most of us1 ,i imagine1 have on occasions come across a ,solitaire board--a round polished board with holes cut in it in a geometrical pattern1 and a glass marble in every hole4 ,sometimes ,i have noticed one on a side table in a suburban front parlour1 or found one on a shelf in a country cottage1 or had one brought under my notice at a wayside inn4 ,sometimes they are of the form shown above1 but it is equally common for the board to have four more holes1 at the points indicated by dots4 ,i select the simpler form4 ,though ,solitaire boards are still sold at the toy shops1 it will be sufficient if the reader will make an enlarged copy of the above on a sheet of cardboard or paper1 number the holes1 and provide himself with #cc counters1 buttons1 or beans4 ,now place a counter in every hole except the central one1 ,no4 #ag1 and the puzzle is to take off all the counters in a series of jumps1 except the last counter1 which must be left in that central hole4 ,you are allowed to jump one counter over the next one to a vacant hole beyond1 just as in the game of draughts1 and the counter jumped over is immediately taken off the board4 ,only remember every move must be a jump2 consequently you will take off a counter at each move1 and thirty-one single jumps will of course remove all the thirty-one counters4 ,but compound moves are allowed 7as in draughts1 again71 for so long as one counter continues to jump1 the jumps all count as one move4 ,here is the beginning of an imaginary solution which will serve to make the manner of moving perfectly plain1 and show how the solver should write out his attempts3 #e-ag1 #ab-aj1 #bf-ab1 #bd-bf 7#ac-aa1 #aa-be71 #i-aa 7#bf-bd1 #bd-aj1 #aj-ab71 etc41 etc4 ,the jumps contained within brackets count as one move1 because they are made with the same counter4 ,find the fewest possible moves4 ,of course1 no diagonal jumps are permitted2 you can only jump in the direction of the lines4 #bbh4--,,the ,,ten ,,apples4 ,illustration ,the family represented in the illustration are amusing themselves with this little puzzle1 which is not very difficult but quite interesting4 ,they have1 it will be seen1 placed sixteen plates on the table in the form of a square1 and put an apple in each of ten plates4 ,they want to find a way of removing all the apples except one by jumping over one at a time to the next vacant square1 as in draughts2 or1 better1 as in solitaire1 for you are not allowed to make any diagonal moves--only moves parallel to the sides of the square4 ,it is obvious that as the apples stand no move can be made1 but you are permitted to transfer any single apple you like to a vacant plate before starting4 ,then the moves must be all leaps1 taking off the apples leaped over4 #bbi4--,,the ,,nine ,,almonds4 ,here is a little puzzle1 said a ,parson1 that ,i have found peculiarly fascinating4 ,it is so simple1 and yet it keeps you interested indefinitely4 ,the reverend gentleman took a sheet of paper and divided it off into twenty-five squares1 like a square portion of a chessboard4 ,then he placed nine almonds on the central squares1 as shown in the illustration1 where we have represented numbered counters for convenience in giving the solution4 ,now1 the puzzle is1 continued the ,parson1 to remove eight of the almonds and leave the ninth in the central square4 ,you make the removals by jumping one almond over another to the vacant square beyond and taking off the one jumped over--just as in draughts1 only here you can jump in any direction1 and not diagonally only4 ,the point is to do the thing in the fewest possible moves4 ,the following specimen attempt will make everything clear4 ,jump #d over #a1 #e over #i1 #c over #f1 #e over #c1 #g over #e and #b1 #d over #g1 #h over #d4 ,but #h is not left in the central square1 as it should be4 ,remember to remove those you jump over4 ,any number of jumps in succession with the same almond count as one move4 ,illustration #bcj4--,,the ,,twelve ,,pennies4 ,here is a pretty little puzzle that only requires twelve pennies or counters4 ,arrange them in a circle1 as shown in the illustration4 ,now take up one penny at a time and1 passing it over two pennies1 place it on the third penny4 ,then take up another single penny and do the same thing1 and so on1 until1 in six such moves1 you have the coins in six pairs in the positions #a1 #b1 #c1 #d1 #e1 #f4 ,you can move in either direction round the circle at every play1 and it does not matter whether the two jumped over are separate or a pair4 ,this is quite easy if you use just a little thought4 ,illustration #bca4--,,plates ,,and ,,coins4 ,place twelve plates1 as shown1 on a round table1 with a penny or orange in every plate4 ,start from any plate you like and1 always going in one direction round the table1 take up one penny1 pass it over two other pennies1 and place it in the next plate4 ,go on again2 take up another penny and1 having passed it over two pennies1 place it in a plate2 and so continue your journey4 ,six coins only are to be removed1 and when these have been placed there should be two coins in each of six plates and six plates empty4 ,an important point of the puzzle is to go round the table as few times as possible4 ,it does not matter whether the two coins passed over are in one or two plates1 nor how many empty plates you pass a coin over4 ,but you must always go in one direction round the table and end at the point from which you set out4 ,your hand1 that is to say1 goes steadily forward in one direction1 without ever moving backwards4 ,illustration #bcb4--;,,catching ,,the ,,mice4 ,illustration ,play fair6 said the mice4 ,you know the rules of the game4 ,yes1 ,i know the rules1 said the cat4 ,i've got to go round and round the circle1 in the direction that you are looking1 and eat every thirteenth mouse1 but ,i must keep the white mouse for a tit-bit at the finish4 ,thirteen is an unlucky number1 but ,i will do my best to oblige you4 ,hurry up1 then6 shouted the mice4 ,give a fellow time to think1 said the cat4 ,i don't know which of you to start at4 ,i must figure it out4 ,while the cat was working out the puzzle he fell asleep1 and1 the spell being thus broken1 the mice returned home in safety4 ,at which mouse should the cat have started the count in order that the white mouse should be the last eaten8 ,when the reader has solved that little puzzle1 here is a second one for him4 ,what is the smallest number that the cat can count round and round the circle1 if he must start at the white mouse 7calling that one in the count7 and still eat the white mouse last of all8 ,and as a third puzzle try to discover what is the smallest number that the cat can count round and round if she must start at the white mouse 7calling that one7 and make the white mouse the third eaten4 #bcc4--,,the ,,eccentric ,,cheesemonger4 ,illustration ,the cheesemonger depicted in the illustration is an inveterate puzzle lover4 ,one of his favourite puzzles is the piling of cheeses in his warehouse1 an amusement that he finds good exercise for the body as well as for the mind4 ,he places sixteen cheeses on the floor in a straight row and then makes them into four piles1 with four cheeses in every pile1 by always passing a cheese over four others4 ,if you use sixteen counters and number them in order from #a to #af1 then you may place #a on #f1 #aa on #a1 #g on #d1 and so on1 until there are four in every pile4 ,it will be seen that it does not matter whether the four passed over are standing alone or piled2 they count just the same1 and you can always carry a cheese in either direction4 ,there are a great many different ways of doing it in twelve moves1 so it makes a good game of patience to try to solve it so that the four piles shall be left in different stipulated places4 ,for example1 try to leave the piles at the extreme ends of the row1 on ,nos4 #a1 #b1 #ae and #af2 this is quite easy4 ,then try to leave three piles together1 on ,nos4 #ac1 #ad1 and #ae4 ,then again play so that they shall be left on ,nos4 #c1 #e1 #ab1 and #ad4 #bcd4--,,the ,,exchange ,,puzzle4 ,here is a rather entertaining little puzzle with moving counters4 ,you only need twelve counters--six of one colour1 marked ,a1 ,c1 ,e1 ,g1 ,i1 and ,k1 and the other six marked ,b1 ,d1 ,f1 ,h1 ,j1 and ,l4 ,you first place them on the diagram1 as shown in the illustration1 and the puzzle is to get them into regular alphabetical order1 as follows3-- ,a ,b ,c ,d ,e ,f ,g ,h ,i ,j ,k ,l ,the moves are made by exchanges of opposite colours standing on the same line4 ,thus1 ,g and ,j may exchange places1 or ,f and ,a1 but you cannot exchange ,g and ,c1 or ,f and ,d1 because in one case they are both white and in the other case both black4 ,can you bring about the required arrangement in seventeen exchanges8 ,illustration ,it cannot be done in fewer moves4 ,the puzzle is really much easier than it looks1 if properly attacked4 #bce4--,,torpedo ,,practice4 ,illustration ,if a fleet of sixteen men-of-war were lying at anchor and surrounded by the enemy1 how many ships might be sunk if every torpedo1 projected in a straight line1 passed under three vessels and sank the fourth8 ,in the diagram we have arranged the fleet in square formation1 where it will be seen that as many as seven ships may be sunk 7those in the top row and first column7 by firing the torpedoes indicated by arrows4 ,anchoring the fleet as we like1 to what extent can we increase this number8 ,remember that each successive ship is sunk before another torpedo is launched1 and that every torpedo proceeds in a different direction2 otherwise1 by placing the ships in a straight line1 we might sink as many as thirteen6 ,it is an interesting little study in naval warfare1 and eminently practical--provided the enemy will allow you to arrange his fleet for your convenience and promise to lie still and do nothing6 #bcf4--,,the ,,hat ,,puzzle4 ,ten hats were hung on pegs as shown in the illustration--five silk hats and five felt bowlers1 alternately silk and felt4 ,the two pegs at the end of the row were empty4 ,illustration ,the puzzle is to remove two contiguous hats to the vacant pegs1 then two other adjoining hats to the pegs now unoccupied1 and so on until five pairs have been moved and the hats again hang in an unbroken row1 but with all the silk ones together and all the felt hats together4 ,remember1 the two hats removed must always be contiguous ones1 and you must take one in each hand and place them on their new pegs without reversing their relative position4 ,you are not allowed to cross your hands1 nor to hang up one at a time4 ,can you solve this old puzzle1 which ,i give as introductory to the next8 ,try it with counters of two colours or with coins1 and remember that the two empty pegs must be left at one end of the row4 #bcg4--;,,boys ,,and ,,girls4 ,if you mark off ten divisions on a sheet of paper to represent the chairs1 and use eight numbered counters for the children1 you will have a fascinating pastime4 ,let the odd numbers represent boys and even numbers girls1 or you can use counters of two colours1 or coins4 ,the puzzle is to remove two children who are occupying adjoining chairs and place them in two empty chairs1 making them first change sides2 then remove a second pair of children from adjoining chairs and place them in the two now vacant1 making them change sides2 and so on1 until all the boys are together and all the girls together1 with the two vacant chairs at one end as at present4 ,to solve the puzzle you must do this in five moves4 ,the two children must always be taken from chairs that are next to one another2 and remember the important point of making the two children change sides1 as this latter is the distinctive feature of the puzzle4 ,by change sides ,i simply mean that if1 for example1 you first move #a and #b to the vacant chairs1 then the first 7the outside7 chair will be occupied by #b and the second one by #a4 ,illustration #bch4--;,,arranging ,,the ,,jampots4 ,i happened to see a little girl sorting out some jam in a cupboard for her mother4 ,she was putting each different kind of preserve apart on the shelves4 ,i noticed that she took a pot of damson in one hand and a pot of gooseberry in the other and made them change places2 then she changed a strawberry with a raspberry1 and so on4 ,it was interesting to observe what a lot of unnecessary trouble she gave herself by making more interchanges than there was any need for1 and ,i thought it would work into a good puzzle4 ,it will be seen in the illustration that little ,dorothy has to manipulate twenty-four large jampots in as many pigeon-holes4 ,she wants to get them in correct numerical order--that is1 #a1 #b1 #c1 #d1 #e1 #f on the top shelf1 #g1 #h1 #i1 #aj1 #aa1 #ab on the next shelf1 and so on4 ,now1 if she always takes one pot in the right hand and another in the left and makes them change places1 how many of these interchanges will be necessary to get all the jampots in proper order8 ,she would naturally first change the #a and the #c1 then the #b and the #c1 when she would have the first three pots in their places4 ,how would you advise her to go on then8 ,place some numbered counters on a sheet of paper divided into squares for the pigeon-holes1 and you will find it an amusing puzzle4 ,illustration ,,unicursal ,,and ,,route ,,problems4 ,i see them on their winding way4 ,,reginald ,,heber4 ,it is reasonable to suppose that from the earliest ages one man has asked another such questions as these3 ,which is the nearest way home8 ,which is the easiest or pleasantest way8 ,how can we find a way that will enable us to dodge the mastodon and the plesiosaurus8 ,how can we get there without ever crossing the track of the enemy8 ,all these are elementary route problems1 and they can be turned into good puzzles by the introduction of some conditions that complicate matters4 ,a variety of such complications will be found in the following examples4 ,i have also included some enumerations of more or less difficulty4 ,these afford excellent practice for the reasoning faculties1 and enable one to generalize in the case of symmetrical forms in a manner that is most instructive4 #bci4--;,a ,,juvenile ,,puzzle4 ,for years ,i have been perpetually consulted by my juvenile friends about this little puzzle4 ,most children seem to know it1 and yet1 curiously enough1 they are invariably unacquainted with the answer4 ,the question they always ask is1 ,do1 please1 tell me whether it is really possible4 ,i believe ,houdin the conjurer used to be very fond of giving it to his child friends1 but ,i cannot say whether he invented the little puzzle or not4 ,no doubt a large number of my readers will be glad to have the mystery of the solution cleared up1 so ,i make no apology for introducing this old teaser4 ,the puzzle is to draw with three strokes of the pencil the diagram that the little girl is exhibiting in the illustration4 ,of course1 you must not remove your pencil from the paper during a stroke or go over the same line a second time4 ,you will find that you can get in a good deal of the figure with one continuous stroke1 but it will always appear as if four strokes are necessary4 ,illustration ,another form of the puzzle is to draw the diagram on a slate and then rub it out in three rubs4 #bdj4--,,the ,,union ,,jack4 ,illustration ,the illustration is a rough sketch somewhat resembling the ,british flag1 the ,union ,jack4 ,it is not possible to draw the whole of it without lifting the pencil from the paper or going over the same line twice4 ,the puzzle is to find out just how much of the drawing it is possible to make without lifting your pencil or going twice over the same line4 ,take your pencil and see what is the best you can do4 #bda4--,,the ,,dissected ,,circle4 ,how many continuous strokes1 without lifting your pencil from the paper1 do you require to draw the design shown in our illustration8 ,directly you change the direction of your pencil it begins a new stroke4 ,you may go over the same line more than once if you like4 ,it requires just a little care1 or you may find yourself beaten by one stroke4 ,illustration #bdb4--,,the ,,tube ,,inspector',s ,,puzzle4 ,the man in our illustration is in a little dilemma4 ,he has just been appointed inspector of a certain system of tube railways1 and it is his duty to inspect regularly1 within a stated period1 all the company's seventeen lines connecting twelve stations1 as shown on the big poster plan that he is contemplating4 ,now he wants to arrange his route so that it shall take him over all the lines with as little travelling as possible4 ,he may begin where he likes and end where he likes4 ,what is his shortest route8 ,could anything be simpler8 ,but the reader will soon find that1 however he decides to proceed1 the inspector must go over some of the lines more than once4 ,in other words1 if we say that the stations are a mile apart1 he will have to travel more than seventeen miles to inspect every line4 ,there is the little difficulty4 ,how far is he compelled to travel1 and which route do you recommend8 ,illustration #bdc4--,,visiting ,,the ,,towns4 ,illustration ,a traveller1 starting from town ,no4 #a1 wishes to visit every one of the towns once1 and once only1 going only by roads indicated by straight lines4 ,how many different routes are there from which he can select8 ,of course1 he must end his journey at ,no4 #a1 from which he started1 and must take no notice of cross roads1 but go straight from town to town4 ,this is an absurdly easy puzzle1 if you go the right way to work4 #bdd4--,,the ,,fifteen ,,turnings4 ,here is another queer travelling puzzle1 the solution of which calls for ingenuity4 ,in this case the traveller starts from the black town and wishes to go as far as possible while making only fifteen turnings and never going along the same road twice4 ,the towns are supposed to be a mile apart4 ,supposing1 for example1 that he went straight to ,a1 then straight to ,b1 then to ,c1 ,d1 ,e1 and ,f1 you will then find that he has travelled thirty-seven miles in five turnings4 ,now1 how far can he go in fifteen turnings8 ,illustration #bde4--,,the ,,fly ,,on ,,the ,,octahedron4 ,look here1 said the professor to his colleague1 ,i have been watching that fly on the octahedron1 and it confines its walks entirely to the edges4 ,what can be its reason for avoiding the sides8 ,perhaps it is trying to solve some route problem1 suggested the other4 ,supposing it to start from the top point1 how many different routes are there by which it may walk over all the edges1 without ever going twice along the same edge in any route8 ,illustration ,the problem was a harder one than they expected1 and after working at it during leisure moments for several days their results did not agree--in fact1 they were both wrong4 ,if the reader is surprised at their failure1 let him attempt the little puzzle himself4 ,i will just explain that the octahedron is one of the five regular1 or ,platonic1 bodies1 and is contained under eight equal and equilateral triangles4 ,if you cut out the two pieces of cardboard of the shape shown in the margin of the illustration1 cut half through along the dotted lines and then bend them and put them together1 you will have a perfect octahedron4 ,in any route over all the edges it will be found that the fly must end at the point of departure at the top4 #bdf4--,,the ,,icosahedron ,,puzzle4 ,the icosahedron is another of the five regular1 or ,platonic1 bodies having all their sides1 angles1 and planes similar and equal4 ,it is bounded by twenty similar equilateral triangles4 ,if you cut out a piece of cardboard of the form shown in the smaller diagram1 and cut half through along the dotted lines1 it will fold up and form a perfect icosahedron4 ,now1 a ,platonic body does not mean a heavenly body2 but it will suit the purpose of our puzzle if we suppose there to be a habitable planet of this shape4 ,we will also suppose that1 owing to a superfluity of water1 the only dry land is along the edges1 and that the inhabitants have no knowledge of navigation4 ,if every one of those edges is #aj1jjj miles long and a solitary traveller is placed at the ,north ,pole 7the highest point shown71 how far will he have to travel before he will have visited every habitable part of the planet--that is1 have traversed every one of the edges8 ,illustration #bdg4--;,,inspecting ,a ,,mine4 ,the diagram is supposed to represent the passages or galleries in a mine4 ,we will assume that every passage1 ,a to ,b1 ,b to ,c1 ,c to ,h1 ,h to ,i1 and so on1 is one furlong in length4 ,it will be seen that there are thirty-one of these passages4 ,now1 an official has to inspect all of them1 and he descends by the shaft to the point ,a4 ,how far must he travel1 and what route do you recommend8 ,the reader may at first say1 ,as there are thirty-one passages1 each a furlong in length1 he will have to travel just thirty-one furlongs4 ,but this is assuming that he need never go along a passage more than once1 which is not the case4 ,take your pencil and try to find the shortest route4 ,you will soon discover that there is room for considerable judgment4 ,in fact1 it is a perplexing puzzle4 ,illustration #bdh4--,,the ,,cyclists' ,,tour4 ,two cyclists were consulting a road map in preparation for a little tour together4 ,the circles represent towns1 and all the good roads are represented by lines4 ,they are starting from the town with a star1 and must complete their tour at ,e4 ,but before arriving there they want to visit every other town once1 and only once4 ,that is the difficulty4 ,mr4 ,spicer said1 ,i am certain we can find a way of doing it2 but ,mr4 ,maggs replied1 ,no way1 ,i'm sure4 ,now1 which of them was correct8 ,take your pencil and see if you can find any way of doing it4 ,of course you must keep to the roads indicated4 ,illustration #bdi4--,,the ,,sailor',s ,,puzzle4 ,the sailor depicted in the illustration stated that he had since his boyhood been engaged in trading with a small vessel among some twenty little islands in the ,pacific4 ,he supplied the rough chart of which ,i have given a copy1 and explained that the lines from island to island represented the only routes that he ever adopted4 ,he always started from island ,a at the beginning of the season1 and then visited every island once1 and once only1 finishing up his tour at the starting-point ,a4 ,but he always put off his visit to ,c as long as possible1 for trade reasons that ,i need not enter into4 ,the puzzle is to discover his exact route1 and this can be done with certainty4 ,take your pencil and1 starting at ,a1 try to trace it out4 ,if you write down the islands in the order in which you visit them--thus1 for example1 ,a1 ,i1 ,o1 ,l1 ,g1 etc4--you can at once see if you have visited an island twice or omitted any4 ,of course1 the crossings of the lines must be ignored--that is1 you must continue your route direct1 and you are not allowed to switch off at a crossing and proceed in another direction4 ,there is no trick of this kind in the puzzle4 ,the sailor knew the best route4 ,can you find it8 ,illustration #bej4--,,the ,,grand ,,tour4 ,one of the everyday puzzles of life is the working out of routes4 ,if you are taking a holiday on your bicycle1 or a motor tour1 there always arises the question of how you are to make the best of your time and other resources4 ,you have determined to get as far as some particular place1 to include visits to such-and-such a town1 to try to see something of special interest elsewhere1 and perhaps to try to look up an old friend at a spot that will not take you much out of your way4 ,then you have to plan your route so as to avoid bad roads1 uninteresting country1 and1 if possible1 the necessity of a return by the same way that you went4 ,with a map before you1 the interesting puzzle is attacked and solved4 ,i will present a little poser based on these lines4 ,i give a rough map of a country--it is not necessary to say what particular country--the circles representing towns and the dotted lines the railways connecting them4 ,now there lived in the town marked ,a a man who was born there1 and during the whole of his life had never once left his native place4 ,from his youth upwards he had been very industrious1 sticking incessantly to his trade1 and had no desire whatever to roam abroad4 ,however1 on attaining his fiftieth birthday he decided to see something of his country1 and especially to pay a visit to a very old friend living at the town marked ,z4 ,what he proposed was this3 that he would start from his home1 enter every town once and only once1 and finish his journey at ,z4 ,as he made up his mind to perform this grand tour by rail only1 he found it rather a puzzle to work out his route1 but he at length succeeded in doing so4 ,how did he manage it8 ,do not forget that every town has to be visited once1 and not more than once4 ,illustration #bea4--,,water1 ,,gas1 ,,and ,,electricity4 ,there are some half-dozen puzzles1 as old as the hills1 that are perpetually cropping up1 and there is hardly a month in the year that does not bring inquiries as to their solution4 ,occasionally one of these1 that one had thought was an extinct volcano1 bursts into eruption in a surprising manner4 ,i have received an extraordinary number of letters respecting the ancient puzzle that ,i have called ,water1 ,gas1 and ,electricity4 ,it is much older than electric lighting1 or even gas1 but the new dress brings it up to date4 ,the puzzle is to lay on water1 gas1 and electricity1 from ,w1 ,g1 and ,e1 to each of the three houses1 ,a1 ,b1 and ,c1 without any pipe crossing another4 ,take your pencil and draw lines showing how this should be done4 ,you will soon find yourself landed in difficulties4 ,illustration #beb4--;,a ,,puzzle ,,for ,,motorists4 ,illustration ,eight motorists drove to church one morning4 ,their respective houses and churches1 together with the only roads available 7the dotted lines71 are shown4 ,one went from his house ,a to his church ,a1 another from his house ,b to his church ,b1 another from ,c to ,c1 and so on1 but it was afterwards found that no driver ever crossed the track of another car4 ,take your pencil and try to trace out their various routes4 #bec4--;,a ,,bank ,,holiday ,,puzzle4 ,two friends were spending their bank holiday on a cycling trip4 ,stopping for a rest at a village inn1 they consulted a route map1 which is represented in our illustration in an exceedingly simplified form1 for the puzzle is interesting enough without all the original complexities4 ,they started from the town in the top left-hand corner marked ,a4 ,it will be seen that there are one hundred and twenty such towns1 all connected by straight roads4 ,now they discovered that there are exactly #a1cfe different routes by which they may reach their destination1 always travelling either due south or due east4 ,the puzzle is to discover which town is their destination4 ,illustration ,of course1 if you find that there are more than #a1cfe different routes to a town it cannot be the right one4 #bed4--,,the ,,motor-,,car ,,tour4 ,illustration ,in the above diagram the circles represent towns and the lines good roads4 ,in just how many different ways can a motorist1 starting from ,london 7marked with an ,l71 make a tour of all these towns1 visiting every town once1 and only once1 on a tour1 and always coming back to ,london on the last ride8 ,the exact reverse of any route is not counted as different4 #bee4--,,the ,,level ,,puzzle4 ,illustration ,this is a simple counting puzzle4 ,in how many different ways can you spell out the word ,,level by placing the point of your pencil on an ,l and then passing along the lines from letter to letter4 ,you may go in any direction1 backwards or forwards4 ,of course you are not allowed to miss letters--that is to say1 if you come to a letter you must use it4 #bef4--,,the ,,diamond ,,puzzle4 ,,in how many different ways may the word ,,diamond be read in the arrangement shown8 ,you may start wherever you like at a ,d and go up or down1 backwards or forwards1 in and out1 in any direction you like1 so long as you always pass from one letter to another that adjoins it4 ,how many ways are there8 ,illustration #beg4--,,the ,,deified ,,puzzle4 ,in how many different ways may the word ,,deified be read in this arrangement under the same conditions as in the last puzzle1 with the addition that you can use any letters twice in the same reading8 ,illustration #beh4--,,the ,,voters' ,,puzzle4 ,illustration ,here we have1 perhaps1 the most interesting form of the puzzle4 ,in how many different ways can you read the political injunction1 ,,rise ,,to ,,vote1 ,,sir1 under the same conditions as before8 ,in this case every reading of the palindrome requires the use of the central ,v as the middle letter4 #bei4--;,,hannah',s ,,puzzle4 ,a man was in love with a young lady whose ,christian name was ,hannah4 ,when he asked her to be his wife she wrote down the letters of her name in this manner3-- ,h ,h ,h ,h ,h ,h ,h ,a ,a ,a ,a ,h ,h ,a ,n ,n ,a ,h ,h ,a ,n ,n ,a ,h ,h ,a ,a ,a ,a ,h ,h ,h ,h ,h ,h ,h and promised that she would be his if he could tell her correctly in how many different ways it was possible to spell out her name1 always passing from one letter to another that was adjacent4 ,diagonal steps are here allowed4 ,whether she did this merely to tease him or to test his cleverness is not recorded1 but it is satisfactory to know that he succeeded4 ,would you have been equally successful8 ,take your pencil and try4 ,you may start from any of the ,h's and go backwards or forwards and in any direction1 so long as all the letters in a spelling are adjoining one another4 ,how many ways are there1 no two exactly alike8 #bfj4--,,the ,,honeycomb ,,puzzle4 ,illustration ,here is a little puzzle with the simplest possible conditions4 ,place the point of your pencil on a letter in one of the cells of the honeycomb1 and trace out a very familiar proverb by passing always from a cell to one that is contiguous to it4 ,if you take the right route you will have visited every cell once1 and only once4 ,the puzzle is much easier than it looks4 #bfa4--,,the ,,monk ,,and ,,the ,,bridges4 ,in this case ,i give a rough plan of a river with an island and five bridges4 ,on one side of the river is a monastery1 and on the other side is seen a monk in the foreground4 ,now1 the monk has decided that he will cross every bridge once1 and only once1 on his return to the monastery4 ,this is1 of course1 quite easy to do1 but on the way he thought to himself1 ,i wonder how many different routes there are from which ,i might have selected4 ,could you have told him8 ,that is the puzzle4 ,take your pencil and trace out a route that will take you once over all the five bridges4 ,then trace out a second route1 then a third1 and see if you can count all the variations4 ,you will find that the difficulty is twofold3 you have to avoid dropping routes on the one hand and counting the same routes more than once on the other4 ,illustration ,,combination ,,and ,,group ,,problems4 ,a combination and a form indeed4 ,hamlet1 iii4 #d4 ,various puzzles in this class might be termed problems in the geometry of situation1 but their solution really depends on the theory of combinations which1 in its turn1 is derived directly from the theory of permutations4 ,it has seemed convenient to include here certain group puzzles and enumerations that might1 perhaps1 with equal reason have been placed elsewhere2 but readers are again asked not to be too critical about the classification1 which is very difficult and arbitrary4 ,as ,i have included my problem of ,the ,round ,table 7,no4 #bgc71 perhaps a few remarks on another well-known problem of the same class1 known by the ,french as ,la ,problme des ,mnages1 may be interesting4 ,if n married ladies are seated at a round table in any determined order1 in how many different ways may their n husbands be placed so that every man is between two ladies but never next to his own wife8 ,this difficult problem was first solved by ,laisant1 and the method shown in the following table is due to ,moreau3-- #d #j #b #e #c #ac #f #ac #hj #g #hc #egi #h #eib #dgch #i #dhba #dcchg #aj #dcigi #dcigib ,the first column shows the number of married couples4 ,the numbers in the second column are obtained in this way3 #e #c #j - #b #ac2 #f #ac #c #b #hc2 #g #hc #ac - #b #eib2 #h #eib #hc #b #dhba2 and so on4 ,find all the numbers1 except #b1 in the table1 and the method will be evident4 ,it will be noted that the #b is subtracted when the first number 7the number of couples7 is odd1 and added when that number is even4 ,the numbers in the third column are obtained thus3 #ac - #j #ac2 #hc - #c #hj2 #eib - #ac #egi2 #dhba - #hc #dgch2 and so on4 ,the numbers in this last column give the required solutions4 ,thus1 four husbands may be seated in two ways1 five husbands may be placed in thirteen ways1 and six husbands in eighty ways4 ,the following method1 by ,lucas1 will show the remarkable way in which chessboard analysis may be applied to the solution of a circular problem of this kind4 ,divide a square into thirty-six cells1 six by six1 and strike out all the cells in the long diagonal from the bottom left-hand corner to the top right-hand corner1 also the five cells in the diagonal next above it and the cell in the bottom right-hand corner4 ,the answer for six couples will be the same as the number of ways in which you can place six rooks 7not using the cancelled cells7 so that no rook shall ever attack another rook4 ,it will be found that the six rooks may be placed in eighty different ways1 which agrees with the above table4 #bfb4--,,those ,,fifteen ,,sheep4 ,a certain cyclopdia has the following curious problem1 ,i am told3 ,place fifteen sheep in four pens so that there shall be the same number of sheep in each pen4 ,no answer whatever is vouchsafed1 so ,i thought ,i would investigate the matter4 ,i saw that in dealing with apples or bricks the thing would appear to be quite impossible1 since four times any number must be an even number1 while fifteen is an odd number4 ,i thought1 therefore1 that there must be some quality peculiar to the sheep that was not generally known4 ,so ,i decided to interview some farmers on the subject4 ,the first one pointed out that if we put one pen inside another1 like the rings of a target1 and placed all sheep in the smallest pen1 it would be all right4 ,but ,i objected to this1 because you admittedly place all the sheep in one pen1 not in four pens4 ,the second man said that if ,i placed four sheep in each of three pens and three sheep in the last pen 7that is fifteen sheep in all71 and one of the ewes in the last pen had a lamb during the night1 there would be the same number in each pen in the morning4 ,this also failed to satisfy me4 ,illustration ,the third farmer said1 ,i've got four hurdle pens down in one of my fields1 and a small flock of wethers1 so if you will just step down with me ,i will show you how it is done4 ,the illustration depicts my friend as he is about to demonstrate the matter to me4 ,his lucid explanation was evidently that which was in the mind of the writer of the article in the cyclopdia4 ,what was it8 ,can you place those fifteen sheep8 #bfc4--,,king ,,arthur',s ,,knights4 ,king ,arthur sat at the ,round ,table on three successive evenings with his knights--,beleobus1 ,caradoc1 ,driam1 ,eric1 ,floll1 and ,galahad--but on no occasion did any person have as his neighbour one who had before sat next to him4 ,on the first evening they sat in alphabetical order round the table4 ,but afterwards ,king ,arthur arranged the two next sittings so that he might have ,beleobus as near to him as possible and ,galahad as far away from him as could be managed4 ,how did he seat the knights to the best advantage1 remembering that rule that no knight may have the same neighbour twice8 #bfd4--,,the ,,city ,,luncheons4 ,twelve men connected with a large firm in the ,city of ,london sit down to luncheon together every day in the same room4 ,the tables are small ones that only accommodate two persons at the same time4 ,can you show how these twelve men may lunch together on eleven days in pairs1 so that no two of them shall ever sit twice together8 ,we will represent the men by the first twelve letters of the alphabet1 and suppose the first day's pairing to be as follows-- 7,a ,b7 7,c ,d7 7,e ,f7 7,g ,h7 7,i ,j7 7,k ,l74 ,then give any pairing you like for the next day1 say-- 7,a ,c7 7,b ,d7 7,e ,g7 7,f ,h7 7,i ,k7 7,j ,l71 and so on1 until you have completed your eleven lines1 with no pair ever occurring twice4 ,there are a good many different arrangements possible4 ,try to find one of them4 #bfe4--;,a ,,puzzle ,,for ,,card-,,players4 ,twelve members of a club arranged to play bridge together on eleven evenings1 but no player was ever to have the same partner more than once1 or the same opponent more than twice4 ,can you draw up a scheme showing how they may all sit down at three tables every evening8 ,call the twelve players by the first twelve letters of the alphabet and try to group them4 #bff4--;,a ,,tennis ,,tournament4 ,four married couples played a mixed double tennis tournament1 a man and a lady always playing against a man and a lady4 ,but no person ever played with or against any other person more than once4 ,can you show how they all could have played together in the two courts on three successive days8 ,this is a little puzzle of a quite practical kind1 and it is just perplexing enough to be interesting4 #bfg4--,,the ,,wrong ,,hats4 ,one of the most perplexing things ,i have come across lately1 said ,mr4 ,wilson1 is this4 ,eight men had been dining not wisely but too well at a certain ,london restaurant4 ,they were the last to leave1 but not one man was in a condition to identify his own hat4 ,now1 considering that they took their hats at random1 what are the chances that every man took a hat that did not belong to him8 ,the first thing1 said ,mr4 ,waterson1 is to see in how many different ways the eight hats could be taken4 ,that is quite easy1 ,mr4 ,stubbs explained4 ,multiply together the numbers1 #a1 #b1 #c1 #d1 #e1 #f1 #g1 and #h4 ,let me see--half a minute--yes2 there are #dj1cbj different ways4 ,now all you've got to do is to see in how many of these cases no man has his own hat1 said ,mr4 ,waterson4 ,thank you1 ,i'm not taking any1 said ,mr4 ,packhurst4 ,i don't envy the man who attempts the task of writing out all those forty-thousand-odd cases and then picking out the ones he wants4 ,they all agreed that life is not long enough for that sort of amusement2 and as nobody saw any other way of getting at the answer1 the matter was postponed indefinitely4 ,can you solve the puzzle8 #bfh4--,,the ,,peal ,,of ,,bells4 ,a correspondent1 who is apparently much interested in campanology1 asks me how he is to construct what he calls a true and correct peal for four bells4 ,he says that every possible permutation of the four bells must be rung once1 and once only4 ,he adds that no bell must move more than one place at a time1 that no bell must make more than two successive strokes in either the first or the last place1 and that the last change must be able to pass into the first4 ,these fantastic conditions will be found to be observed in the little peal for three bells1 as follows3-- #a #b #c #b #a #c #b #c #a #c #b #a #c #a #b #a #c #b ,how are we to give him a correct solution for his four bells8 #bfi4--,,three ,,men ,,in ,a ,,boat4 ,a certain generous ,london manufacturer gives his workmen every year a week's holiday at the seaside at his own expense4 ,one year fifteen of his men paid a visit to ,herne ,bay4 ,on the morning of their departure from ,london they were addressed by their employer1 who expressed the hope that they would have a very pleasant time4 ,i have been given to understand1 he added1 that some of you fellows are very fond of rowing1 so ,i propose on this occasion to provide you with this recreation1 and at the same time give you an amusing little puzzle to solve4 ,during the seven days that you are at ,herne ,bay every one of you will go out every day at the same time for a row1 but there must always be three men in a boat and no more4 ,no two men may ever go out in a boat together more than once1 and no man is allowed to go out twice in the same boat4 ,if you can manage to do this1 and use as few different boats as possible1 you may charge the firm with the expense4 ,one of the men tells me that the experience he has gained in such matters soon enabled him to work out the answer to the entire satisfaction of themselves and their employer4 ,but the amusing part of the thing is that they never really solved the little mystery4 ,i find their method to have been quite incorrect1 and ,i think it will amuse my readers to discover how the men should have been placed in the boats4 ,as their names happen to have been ,andrews1 ,baker1 ,carter1 ,danby1 ,edwards1 ,frith1 ,gay1 ,hart1 ,isaacs1 ,jackson1 ,kent1 ,lang1 ,mason1 ,napper1 and ,onslow1 we can call them by their initials and write out the five groups for each of the seven days in the following simple way3 #a #b #c #d #e ,first ,day3 7,,abc7 7,,def7 7,,ghi7 7,,jkl7 7,,mno74 ,the men within each pair of brackets are here seen to be in the same boat1 and therefore ,a can never go out with ,b or with ,c again1 and ,c can never go out again with ,b4 ,the same applies to the other four boats4 ,the figures show the number on the boat1 so that ,a1 ,b1 or ,c1 for example1 can never go out in boat ,no4 #a again4 #bgj4--,,the ,,glass ,,balls4 ,a number of clever marksmen were staying at a country house1 and the host1 to provide a little amusement1 suspended strings of glass balls1 as shown in the illustration1 to be fired at4 ,after they had all put their skill to a sufficient test1 somebody asked the following question3 ,what is the total number of different ways in which these sixteen balls may be broken1 if we must always break the lowest ball that remains on any string8 ,thus1 one way would be to break all the four balls on each string in succession1 taking the strings from left to right4 ,another would be to break all the fourth balls on the four strings first1 then break the three remaining on the first string1 then take the balls on the three other strings alternately from right to left1 and so on4 ,there is such a vast number of different ways 7since every little variation of order makes a different way7 that one is apt to be at first impressed by the great difficulty of the problem4 ,yet it is really quite simple when once you have hit on the proper method of attacking it4 ,how many different ways are there8 ,illustration #bga4--;,,fifteen ,,letter ,,puzzle4 ,,ale ,,foe ,,hod ,,bgn ,,cab ,,hen ,,jog ,,kfm ,,hag ,,gem ,,mob ,,bfh ,,fan ,,kin ,,jek ,,dfl ,,jam ,,him ,,gcl ,,ljh ,,aid ,,jib ,,fcj ,,njd ,,oak ,,fig ,,hck ,,mln ,,bed ,,oil ,,mcd ,,blk ,,ice ,,con ,,dgk ,the above is the solution of a puzzle ,i gave in ,tit-bits in the summer of #ahif4 ,it was required to take the letters1 ,a1 ,b1 ,c1 ,d1 ,e1 ,f1 ,g1 ,h1 ,i1 ,j1 ,k1 ,l1 ,m1 ,n1 and ,o1 and with them form thirty-five groups of three letters so that the combinations should include the greatest number possible of common ,english words4 ,no two letters may appear together in a group more than once4 ,thus1 ,a and ,l having been together in ,,ale1 must never be found together again2 nor may ,a appear again in a group with ,e1 nor ,l with ,e4 ,these conditions will be found complied with in the above solution1 and the number of words formed is twenty-one4 ,many persons have since tried hard to beat this number1 but so far have not succeeded4 ,more than thirty-five combinations of the fifteen letters cannot be formed within the conditions4 ,theoretically1 there cannot possibly be more than twenty-three words formed1 because only this number of combinations is possible with a vowel or vowels in each4 ,and as no ,english word can be formed from three of the given vowels 7,a1 ,e1 ,i1 and ,o71 we must reduce the number of possible words to twenty-two4 ,this is correct theoretically1 but practically that twenty-second word cannot be got in4 ,if ,,jek1 shown above1 were a word it would be all right2 but it is not1 and no amount of juggling with the other letters has resulted in a better answer than the one shown4 ,i should1 say that proper nouns and abbreviations1 such as ,joe1 ,jim1 ,alf1 ,hal1 ,flo1 ,ike1 etc41 are disallowed4 ,now1 the present puzzle is a variation of the above4 ,it is simply this3 ,instead of using the fifteen letters given1 the reader is allowed to select any fifteen different letters of the alphabet that he may prefer4 ,then construct thirty-five groups in accordance with the conditions1 and show as many good ,english words as possible4 #bgb4--,,the ,,nine ,,schoolboys4 ,this is a new and interesting companion puzzle to the ,fifteen ,schoolgirls 7see solution of ,no4 #bfi71 and even in the simplest possible form in which ,i present it there are unquestionable difficulties4 ,nine schoolboys walk out in triplets on the six week days so that no boy ever walks side by side with any other boy more than once4 ,how would you arrange them8 ,if we represent them by the first nine letters of the alphabet1 they might be grouped on the first day as follows3-- ,a ,b ,c ,d ,e ,f ,g ,h ,i ,then ,a can never walk again side by side with ,b1 or ,b with ,c1 or ,d with ,e1 and so on4 ,but ,a can1 of course1 walk side by side with ,c4 ,it is here not a question of being together in the same triplet1 but of walking side by side in a triplet4 ,under these conditions they can walk out on six days2 under the ,schoolgirls conditions they can only walk on four days4 #bgc4--,,the ,,round ,,table4 ,seat the same n persons at a round table on 7n - #a77n - #b7 -------------- #b occasions so that no person shall ever have the same two neighbours twice4 ,this is1 of course1 equivalent to saying that every person must sit once1 and once only1 between every possible pair4 #bgd4--,,the ,,mouse-,,trap ,,puzzle4 ,illustration #f #bj #b #ai #ac #ba #g #e #c #ah #ag #h #ae #aa #ad #af #a #i #aj #d #ab ,this is a modern version1 with a difference1 of an old puzzle of the same name4 ,number twenty-one cards1 #a1 #b1 #c1 etc41 up to #ba1 and place them in a circle in the particular order shown in the illustration4 ,these cards represent mice4 ,you start from any card1 calling that card one1 and count1 one1 two1 three1 etc41 in a clockwise direction1 and when your count agrees with the number on the card1 you have made a catch1 and you remove the card4 ,then start at the next card1 calling that one1 and try again to make another catch4 ,and so on4 ,supposing you start at #ah1 calling that card one1 your first catch will be #ai4 ,remove #ai and your next catch is #aj4 ,remove #aj and your next catch is #a4 ,remove the #a1 and if you count up to #ba 7you must never go beyond71 you cannot make another catch4 ,now1 the ideal is to catch all the twenty-one mice1 but this is not here possible1 and if it were it would merely require twenty-one different trials1 at the most1 to succeed4 ,but the reader may make any two cards change places before he begins4 ,thus1 you can change the #f with the #b1 or the #g with the #aa1 or any other pair4 ,this can be done in several ways so as to enable you to catch all the twenty-one mice1 if you then start at the right place4 ,you may never pass over a catch2 you must always remove the card and start afresh4 #bge4--,,the ,,sixteen ,,sheep4 ,illustration3 #j #j #j #j --------------------- #j #j #j #j #j #j #j #j ----------- #j #j #j #j ,here is a new puzzle with matches and counters or coins4 ,in the illustration the matches represent hurdles and the counters sheep4 ,the sixteen hurdles on the outside1 and the sheep1 must be regarded as immovable2 the puzzle has to do entirely with the nine hurdles on the inside4 ,it will be seen that at present these nine hurdles enclose four groups of #h1 #c1 #c1 and #b sheep4 ,the farmer requires to readjust some of the hurdles so as to enclose #f1 #f1 and #d sheep4 ,can you do it by only replacing two hurdles8 ,when you have succeeded1 then try to do it by replacing three hurdles2 then four1 five1 six1 and seven in succession4 ,of course1 the hurdles must be legitimately laid on the dotted lines1 and no such tricks are allowed as leaving unconnected ends of hurdles1 or two hurdles placed side by side1 or merely making hurdles change places4 ,in fact1 the conditions are so simple that any farm labourer will understand it directly4 #bgf4--,,the ,,eight ,,villas4 ,in one of the outlying suburbs of ,london a man had a square plot of ground on which he decided to build eight villas1 as shown in the illustration1 with a common recreation ground in the middle4 ,after the houses were completed1 and all or some of them let1 he discovered that the number of occupants in the three houses forming a side of the square was in every case nine4 ,he did not state how the occupants were distributed1 but ,i have shown by the numbers on the sides of the houses one way in which it might have happened4 ,the puzzle is to discover the total number of ways in which all or any of the houses might be occupied1 so that there should be nine persons on each side4 ,in order that there may be no misunderstanding1 ,i will explain that although ,b is what we call a reflection of ,a1 these would count as two different arrangements1 while ,c1 if it is turned round1 will give four arrangements2 and if turned round in front of a mirror1 four other arrangements4 ,all eight must be counted4 ,illustration3 _/ _/ _/ #b #e #b _/ _/ #e #e _/ _/ _/ #b #e #b --------- --------- --------- #a #f #b #b #f #a #a #f #b --------- --------- --------- #f #f #f #f #d #d --------- --------- --------- #b #f #a #a #f #b #d #b #c --------- --------- --------- ,a ,b ,c #bgg4--;,,counter ,,crosses4 ,all that we need for this puzzle is nine counters1 numbered #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 and #i4 ,it will be seen that in the illustration ,a these are arranged so as to form a ,greek cross1 while in the case of ,b they form a ,latin cross4 ,in both cases the reader will find that the sum of the numbers in the upright of the cross is the same as the sum of the numbers in the horizontal arm4 ,it is quite easy to hit on such an arrangement by trial1 but the problem is to discover in exactly how many different ways it may be done in each case4 ,remember that reversals and reflections do not count as different4 ,that is to say1 if you turn this page round you get four arrangements of the ,greek cross1 and if you turn it round again in front of a mirror you will get four more4 ,but these eight are all regarded as one and the same4 ,now1 how many different ways are there in each case8 ,illustration3 7#a7 7#b7 7#b7 7#d7 7#e7 7#a7 7#f7 7#g7 7#c7 7#d7 7#i7 7#e7 7#f7 7#c7 7#g7 7#h7 ,a 7#h7 ,b 7#i7 #bgh4--;,a ,,dormitory ,,puzzle4 ,in a certain convent there were eight large dormitories on one floor1 approached by a spiral staircase in the centre1 as shown in our plan4 ,on an inspection one ,monday by the abbess it was found that the south aspect was so much preferred that six times as many nuns slept on the south side as on each of the other three sides4 ,she objected to this overcrowding1 and ordered that it should be reduced4 ,on ,tuesday she found that five times as many slept on the south side as on each of the other sides4 ,again she complained4 ,on ,wednesday she found four times as many on the south side1 on ,thursday three times as many1 and on ,friday twice as many4 ,urging the nuns to further efforts1 she was pleased to find on ,saturday that an equal number slept on each of the four sides of the house4 ,what is the smallest number of nuns there could have been1 and how might they have arranged themselves on each of the six nights8 ,no room may ever be unoccupied4 ,illustration --------- --------- _/ -99- _/ --------- --------- #bgi4--,,the ,,barrels ,,of ,,balsam4 ,a merchant of ,bagdad had ten barrels of precious balsam for sale4 ,they were numbered1 and were arranged in two rows1 one on top of the other1 as shown in the picture4 ,the smaller the number on the barrel1 the greater was its value4 ,so that the best quality was numbered #a and the worst numbered #aj1 and all the other numbers of graduating values4 ,now1 the rule of ,ahmed ,assan1 the merchant1 was that he never put a barrel either beneath or to the right of one of less value4 ,the arrangement shown is1 of course1 the simplest way of complying with this condition4 ,but there are many other ways--such1 for example1 as this3-- #a #b #e #g #h #c #d #f #i #aj ,here1 again1 no barrel has a smaller number than itself on its right or beneath it4 ,the puzzle is to discover in how many different ways the merchant of ,bagdad might have arranged his barrels in the two rows without breaking his rule4 ,can you count the number of ways8 #bhj4--;,,building ,,the ,,tetrahedron4 ,i possess a tetrahedron1 or triangular pyramid1 formed of six sticks glued together1 as shown in the illustration4 ,can you count correctly the number of different ways in which these six sticks might have been stuck together so as to form the pyramid8 ,some friends worked at it together one evening1 each person providing himself with six lucifer matches to aid his thoughts2 but it was found that no two results were the same4 ,you see1 if we remove one of the sticks and turn it round the other way1 that will be a different pyramid4 ,if we make two of the sticks change places the result will again be different4 ,but remember that every pyramid may be made to stand on either of its four sides without being a different one4 ,how many ways are there altogether8 ,illustration #bha4--,,painting ,a ,,pyramid4 ,this puzzle concerns the painting of the four sides of a tetrahedron1 or triangular pyramid4 ,if you cut out a piece of cardboard of the triangular shape shown in ,fig4 #a1 and then cut half through along the dotted lines1 it will fold up and form a perfect triangular pyramid4 ,and ,i would first remind my readers that the primary colours of the solar spectrum are seven--violet1 indigo1 blue1 green1 yellow1 orange1 and red4 ,when ,i was a child ,i was taught to remember these by the ungainly word formed by the initials of the colours1 ,vibgyor4 ,illustration ,in how many different ways may the triangular pyramid be coloured1 using in every case one1 two1 three1 or four colours of the solar spectrum8 ,of course a side can only receive a single colour1 and no side can be left uncoloured4 ,but there is one point that ,i must make quite clear4 ,the four sides are not to be regarded as individually distinct4 ,that is to say1 if you paint your pyramid as shown in ,fig4 #b 7where the bottom side is green and the other side that is out of view is yellow71 and then paint another in the order shown in ,fig4 #c1 these are really both the same and count as one way4 ,for if you tilt over ,no4 #b to the right it will so fall as to represent ,no4 #c4 ,the avoidance of repetitions of this kind is the real puzzle of the thing4 ,if a coloured pyramid cannot be placed so that it exactly resembles in its colours and their relative order another pyramid1 then they are different4 ,remember that one way would be to colour all the four sides red1 another to colour two sides green1 and the remaining sides yellow and blue2 and so on4 #bhb4--,,the ,,antiquary',s ,,chain4 ,an antiquary possessed a number of curious old links1 which he took to a blacksmith1 and told him to join together to form one straight piece of chain1 with the sole condition that the two circular links were not to be together4 ,the following illustration shows the appearance of the chain and the form of each link4 ,now1 supposing the owner should separate the links again1 and then take them to another smith and repeat his former instructions exactly1 what are the chances against the links being put together exactly as they were by the first man8 ,remember that every successive link can be joined on to another in one of two ways1 just as you can put a ring on your finger in two ways1 or link your forefingers and thumbs in two ways4 ,illustration #bhc4--,,the ,,fifteen ,,dominoes4 ,in this case we do not use the complete set of twenty-eight dominoes to be found in the ordinary box4 ,we dispense with all those dominoes that have a five or a six on them and limit ourselves to the fifteen that remain1 where the double-four is the highest4 ,in how many different ways may the fifteen dominoes be arranged in a straight line in accordance with the simple rule of the game that a number must always be placed against a similar number--that is1 a four against a four1 a blank against a blank1 and so on8 ,left to right and right to left of the same arrangement are to be counted as two different ways4 #chd4--,,the ,,cross ,,target4 -- 9999 -- 9999 ------ 99 ------ 99 99 ------ 99 -- -- ,in the illustration we have a somewhat curious target designed by an eccentric sharpshooter4 ,his idea was that in order to score you must hit four circles in as many shots so that those four shots shall form a square4 ,it will be seen by the results recorded on the target that two attempts have been successful4 ,the first man hit the four circles at the top of the cross1 and thus formed his square4 ,the second man intended to hit the four in the bottom arm1 but his second shot1 on the left1 went too high4 ,this compelled him to complete his four in a different way than he intended4 ,it will thus be seen that though it is immaterial which circle you hit at the first shot1 the second shot may commit you to a definite procedure if you are to get your square4 ,now1 the puzzle is to say in just how many different ways it is possible to form a square on the target with four shots4 #bhe4--,,the ,,four ,,postage ,,stamps4 --------------- #a #b #c #d --------------- #e #f #g #h --------------- #i #aj #aa #ab --------------- ,it is as easy as counting1 is an expression one sometimes hears4 ,but mere counting may be puzzling at times4 ,take the following simple example4 ,suppose you have just bought twelve postage stamps1 in this form--three by four--and a friend asks you to oblige him with four stamps1 all joined together--no stamp hanging on by a mere corner4 ,in how many different ways is it possible for you to tear off those four stamps8 ,you see1 you can give him #a1 #b1 #c1 #d1 or #b1 #c1 #f1 #g1 or #a1 #b1 #c1 #f1 or #a1 #b1 #c1 #g1 or #b1 #c1 #d1 #h1 and so on4 ,can you count the number of different ways in which those four stamps might be delivered8 ,there are not many more than fifty ways1 so it is not a big count4 ,can you get the exact number8 #bhf4--,,painting ,,the ,,die4 ,in how many different ways may the numbers on a single die be marked1 with the only condition that the #a and #f1 the #b and #e1 and the #c and #d must be on opposite sides8 ,it is a simple enough question1 and yet it will puzzle a good many people4 #bhg4--;,,an ,,acrostic ,,puzzle4 ,in the making or solving of double acrostics1 has it ever occurred to you to consider the variety and limitation of the pair of initial and final letters available for cross words8 ,you may have to find a word beginning with ,a and ending with ,b1 or ,a and ,c1 or ,a and ,d1 and so on4 ,some combinations are obviously impossible--such1 for example1 as those with ,q at the end4 ,but let us assume that a good ,english word can be found for every case4 ,then how many possible pairs of letters are available8 ,,chessboard ,,problems4 ,you and ,i will goe to the chesse4 ,,greene',s ,groatsworth of ,wit4 ,during a heavy gale a chimney-pot was hurled through the air1 and crashed upon the pavement just in front of a pedestrian4 ,he quite calmly said1 ,i have no use for it3 ,i do not smoke4 ,some readers1 when they happen to see a puzzle represented on a chessboard with chess pieces1 are apt to make the equally inconsequent remark1 ,i have no use for it3 ,i do not play chess4 ,this is largely a result of the common1 but erroneous1 notion that the ordinary chess puzzle with which we are familiar in the press 7dignified1 for some reason1 with the name problem7 has a vital connection with the game of chess itself4 ,but there is no condition in the game that you shall checkmate your opponent in two moves1 in three moves1 or in four moves1 while the majority of the positions given in these puzzles are such that one player would have so great a superiority in pieces that the other would have resigned before the situations were reached4 ,and the solving of them helps you but little1 and that quite indirectly1 in playing the game1 it being well known that1 as a rule1 the best chess problemists are indifferent players1 and vice versa4 ,occasionally a man will be found strong on both subjects1 but he is the exception to the rule4 ,yet the simple chequered board and the characteristic moves of the pieces lend themselves in a very remarkable manner to the devising of the most entertaining puzzles4 ,there is room for such infinite variety that the true puzzle lover cannot afford to neglect them4 ,it was with a view to securing the interest of readers who are frightened off by the mere presentation of a chessboard that so many puzzles of this class were originally published by me in various fanciful dresses4 ,some of these posers ,i still retain in their disguised form2 others ,i have translated into terms of the chessboard4 ,in the majority of cases the reader will not need any knowledge whatever of chess1 but ,i have thought it best to assume throughout that he is acquainted with the terminology1 the moves1 and the notation of the game4 ,i first deal with a few questions affecting the chessboard itself2 then with certain statical puzzles relating to the ,rook1 the ,bishop1 the ,queen1 and the ,knight in turn2 then dynamical puzzles with the pieces in the same order2 and1 finally1 with some miscellaneous puzzles on the chessboard4 ,it is hoped that the formul and tables given at the end of the statical puzzles will be of interest1 as they are1 for the most part1 published for the first time4 ,,the ,,chessboard4 ,good company's a chessboard4 ,,byron',s ,don ,juan1 xiii4 #hi4 ,a chessboard is essentially a square plane divided into sixty-four smaller squares by straight lines at right angles4 ,originally it was not chequered 7that is1 made with its rows and columns alternately black and white1 or of any other two colours71 and this improvement was introduced merely to help the eye in actual play4 ,the utility of the chequers is unquestionable4 ,for example1 it facilitates the operation of the bishops1 enabling us to see at the merest glance that our king or pawns on black squares are not open to attack from an opponent's bishop running on the white diagonals4 ,yet the chequering of the board is not essential to the game of chess4 ,also1 when we are propounding puzzles on the chessboard1 it is often well to remember that additional interest may result from generalizing for boards containing any number of squares1 or from limiting ourselves to some particular chequered arrangement1 not necessarily a square4 ,we will give a few puzzles dealing with chequered boards in this general way4 #bhh4--;,,chequered ,,board ,,divisions4 ,i recently asked myself the question3 ,in how many different ways may a chessboard be divided into two parts of the same size and shape by cuts along the lines dividing the squares8 ,the problem soon proved to be both fascinating and bristling with difficulties4 ,i present it in a simplified form1 taking a board of smaller dimensions4 ,illustration3 ------99------ ---------99--- ---------99--- ,h ,h ,h ------99------ ------9999--- ---9999---99--- ,h ,h ,h ,h ,h ------99------ ------99------ ---99---99---99--- ,h ,h ,h ,h ,h ------99------ ---9999------ ---99---9999--- ,h ,h ,h ------99------ ---99--------- ---99--------- ------------------ ------------------ ------------------ ------------------ ------------------ ------------------ ------------------ ------99------ ---------99--- ---------99--- ,h ,h ,h ---9999------ ---999999--- ------9999--- ,h ,h ,h ---999999--- ---999999--- ------99------ ,h ,h ,h ------9999--- ---999999--- ---9999------ ,h ,h ,h ------99------ ---99--------- ---99--------- ,it is obvious that a board of four squares can only be so divided in one way--by a straight cut down the centre--because we shall not count reversals and reflections as different4 ,in the case of a board of sixteen squares--four by four--there are just six different ways4 ,i have given all these in the diagram1 and the reader will not find any others4 ,now1 take the larger board of thirty-six squares1 and try to discover in how many ways it may be cut into two parts of the same size and shape4 #bhi4--,,lions ,,and ,,crowns4 ,the young lady in the illustration is confronted with a little cutting-out difficulty in which the reader may be glad to assist her4 ,she wishes1 for some reason that she has not communicated to me1 to cut that square piece of valuable material into four parts1 all of exactly the same size and shape1 but it is important that every piece shall contain a lion and a crown4 ,as she insists that the cuts can only be made along the lines dividing the squares1 she is considerably perplexed to find out how it is to be done4 ,can you show her the way8 ,there is only one possible method of cutting the stuff4 ,illustration3 ------ ------ ,l,l,l ------ ,c,c ------ ,c,c ------ ,l ------ ------ #bij4--;,,boards ,,with ,,an ,,odd ,,number ,,of ,,squares4 ,we will here consider the question of those boards that contain an odd number of squares4 ,we will suppose that the central square is first cut out1 so as to leave an even number of squares for division4 ,now1 it is obvious that a square three by three can only be divided in one way1 as shown in ,fig4 #a4 ,it will be seen that the pieces ,a and ,b are of the same size and shape1 and that any other way of cutting would only produce the same shaped pieces1 so remember that these variations are not counted as different ways4 ,the puzzle ,i propose is to cut the board five by five 7,fig4 #b7 into two pieces of the same size and shape in as many different ways as possible4 ,i have shown in the illustration one way of doing it4 ,how many different ways are there altogether8 ,a piece which when turned over resembles another piece is not considered to be of a different shape4 ,illustration3 ---99------ ,h ---9999--- ,,hhhhh ---9999--- ,h ------99--- ,fig #a ,illustration3 --------------- 99999999------ ,h ------9999------ ,,hhhhh ------9999------ ,h ------99999999 ,h ---99------------ ,fig #b #bia4--,,the ,,grand ,,lama',s ,,problem4 ,once upon a time there was a ,grand ,lama who had a chessboard made of pure gold1 magnificently engraved1 and1 of course1 of great value4 ,every year a tournament was held at ,lhassa among the priests1 and whenever any one beat the ,grand ,lama it was considered a great honour1 and his name was inscribed on the back of the board1 and a costly jewel set in the particular square on which the checkmate had been given4 ,after this sovereign pontiff had been defeated on four occasions he died--possibly of chagrin4 ,illustration3 ------------------------ 99 ------------------------ 99 ------------------------ 99 ------------------------ 99 ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ,now the new ,grand ,lama was an inferior chess-player1 and preferred other forms of innocent amusement1 such as cutting off people's heads4 ,so he discouraged chess as a degrading game1 that did not improve either the mind or the morals1 and abolished the tournament summarily4 ,then he sent for the four priests who had had the effrontery to play better than a ,grand ,lama1 and addressed them as follows3 ,miserable and heathenish men1 calling yourselves priests6 ,know ye not that to lay claim to a capacity to do anything better than my predecessor is a capital offence8 ,take that chessboard and1 before day dawns upon the torture chamber1 cut it into four equal parts of the same shape1 each containing sixteen perfect squares1 with one of the gems in each part6 ,if in this you fail1 then shall other sports be devised for your special delectation4 ,go6 ,the four priests succeeded in their apparently hopeless task4 ,can you show how the board may be divided into four equal parts1 each of exactly the same shape1 by cuts along the lines dividing the squares1 each part to contain one of the gems8 #bib4--,,the ,,abbot',s ,,window4 ,illustration ,once upon a time the ,lord ,abbot of ,st4 ,edmondsbury1 in consequence of devotions too strong for his head1 fell sick and was unable to leave his bed4 ,as he lay awake1 tossing his head restlessly from side to side1 the attentive monks noticed that something was disturbing his mind2 but nobody dared ask what it might be1 for the abbot was of a stern disposition1 and never would brook inquisitiveness4 ,suddenly he called for ,father ,john1 and that venerable monk was soon at the bedside4 ,father ,john1 said the ,abbot1 dost thou know that ,i came into this wicked world on a ,christmas ,even8 ,the monk nodded assent4 ,and have ,i not often told thee that1 having been born on ,christmas ,even1 ,i have no love for the things that are odd8 ,look there6 ,the ,abbot pointed to the large dormitory window1 of which ,i give a sketch4 ,the monk looked1 and was perplexed4 ,dost thou not see that the sixty-four lights add up an even number vertically and horizontally1 but that all the diagonal lines1 except fourteen are of a number that is odd8 ,why is this8 ,of a truth1 my ,lord ,abbot1 it is of the very nature of things1 and cannot be changed4 ,nay1 but it shall be changed4 ,i command thee that certain of the lights be closed this day1 so that every line shall have an even number of lights4 ,see thou that this be done without delay1 lest the cellars be locked up for a month and other grievous troubles befall thee4 ,father ,john was at his wits' end1 but after consultation with one who was learned in strange mysteries1 a way was found to satisfy the whim of the ,lord ,abbot4 ,which lights were blocked up1 so that those which remained added up an even number in every line horizontally1 vertically1 and diagonally1 while the least possible obstruction of light was caused8 #bic4--,,the ,,chinese ,,chessboard4 ,into how large a number of different pieces may the chessboard be cut 7by cuts along the lines only71 no two pieces being exactly alike8 ,remember that the arrangement of black and white constitutes a difference4 ,thus1 a single black square will be different from a single white square1 a row of three containing two white squares will differ from a row of three containing two black1 and so on4 ,if two pieces cannot be placed on the table so as to be exactly alike1 they count as different4 ,and as the back of the board is plain1 the pieces cannot be turned over4 #bid4--,,the ,,chessboard ,,sentence4 ,illustration ,i once set myself the amusing task of so dissecting an ordinary chessboard into letters of the alphabet that they would form a complete sentence4 ,it will be seen from the illustration that the pieces assembled give the sentence1 ,,cut ,,thy ,,life1 with the stops between4 ,the ideal sentence would1 of course1 have only one full stop1 but that ,i did not succeed in obtaining4 ,the sentence is an appeal to the transgressor to cut himself adrift from the evil life he is living4 ,can you fit these pieces together to form a perfect chessboard8 ,,statical ,,chess ,,puzzles4 ,they also serve who only stand and wait4 ,,milton4 #bie4--,,the ,,eight ,,rooks4 ,illustration3 ------------------------ ------------------------ ------------------------ ------------------------ ,r ,r ,r ,r ,r ,r ,r ,r ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ,,fig4 #a4 ,illustration3 ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,r ------------------------ ,,fig4 #b4 ,it will be seen in the first diagram that every square on the board is either occupied or attacked by a rook1 and that every rook is guarded 7if they were alternately black and white rooks we should say attacked7 by another rook4 ,placing the eight rooks on any row or file obviously will have the same effect4 ,in diagram #b every square is again either occupied or attacked1 but in this case every rook is unguarded4 ,now1 in how many different ways can you so place the eight rooks on the board that every square shall be occupied or attacked and no rook ever guarded by another8 ,i do not wish to go into the question of reversals and reflections on this occasion1 so that placing the rooks on the other diagonal will count as different1 and similarly with other repetitions obtained by turning the board round4 #bif4--,,the ,,four ,,lions4 ,the puzzle is to find in how many different ways the four lions may be placed so that there shall never be more than one lion in any row or column4 ,mere reversals and reflections will not count as different4 ,thus1 regarding the example given1 if we place the lions in the other diagonal1 it will be considered the same arrangement4 ,for if you hold the second arrangement in front of a mirror or give it a quarter turn1 you merely get the first arrangement4 ,it is a simple little puzzle1 but requires a certain amount of careful consideration4 ,illustration ------------ ,l ------------ ,l ------------ ,l ------------ ,l ------------ #big4--;,,bishops--,,unguarded4 ,place as few bishops as possible on an ordinary chessboard so that every square of the board shall be either occupied or attacked4 ,it will be seen that the rook has more scope than the bishop3 for wherever you place the former1 it will always attack fourteen other squares2 whereas the latter will attack seven1 nine1 eleven1 or thirteen squares1 according to the position of the diagonal on which it is placed4 ,and it is well here to state that when we speak of diagonals in connection with the chessboard1 we do not limit ourselves to the two long diagonals from corner to corner1 but include all the shorter lines that are parallel to these4 ,to prevent misunderstanding on future occasions1 it will be well for the reader to note carefully this fact4 #bih4--;,,bishops--,,guarded4 ,now1 how many bishops are necessary in order that every square shall be either occupied or attacked1 and every bishop guarded by another bishop8 ,and how may they be placed8 #bii4--;,,bishops ,,in ,,convocation4 ,illustration3 ------------------------ ,b ,b ,b ,b ,b ,b ,b ,b ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ,b ,b ,b ,b ,b ,b ------------------------ ,the greatest number of bishops that can be placed at the same time on the chessboard1 without any bishop attacking another1 is fourteen4 ,i show1 in diagram1 the simplest way of doing this4 ,in fact1 on a square chequered board of any number of squares the greatest number of bishops that can be placed without attack is always two less than twice the number of squares on the side4 ,it is an interesting puzzle to discover in just how many different ways the fourteen bishops may be so placed without mutual attack4 ,i shall give an exceedingly simple rule for determining the number of ways for a square chequered board of any number of squares4 #cjj4--,,the ,,eight ,,queens4 ,illustration3 ------------------------ 44,q ---------444------------ 44,q44 ---444------------------ ,q44 ------------------------ ,q ------------------------ ,q ------------------------ 44,q ------------------444--- 44,q44 ------------444--------- ,q44 ------------------------ ,the queen is by far the strongest piece on the chessboard4 ,if you place her on one of the four squares in the centre of the board1 she attacks no fewer than twenty-seven other squares2 and if you try to hide her in a corner1 she still attacks twenty-one squares4 ,eight queens may be placed on the board so that no queen attacks another1 and it is an old puzzle 7first proposed by ,nauck in #ahej1 and it has quite a little literature of its own7 to discover in just how many different ways this may be done4 ,i show one way in the diagram1 and there are in all twelve of these fundamentally different ways4 ,these twelve produce ninety-two ways if we regard reversals and reflections as different4 ,the diagram is in a way a symmetrical arrangement4 ,if you turn the page upside down1 it will reproduce itself exactly2 but if you look at it with one of the other sides at the bottom1 you get another way that is not identical4 ,then if you reflect these two ways in a mirror you get two more ways4 ,now1 all the other eleven solutions are non-symmetrical1 and therefore each of them may be presented in eight ways by these reversals and reflections4 ,it will thus be seen why the twelve fundamentally different solutions produce only ninety-two arrangements1 as ,i have said1 and not ninety-six1 as would happen if all twelve were non-symmetrical4 ,it is well to have a clear understanding on the matter of reversals and reflections when dealing with puzzles on the chessboard4 ,can the reader place the eight queens on the board so that no queen shall attack another and so that no three queens shall be in a straight line in any oblique direction8 ,another glance at the diagram will show that this arrangement will not answer the conditions1 for in the two directions indicated by the dotted lines there are three queens in a straight line4 ,there is only one of the twelve fundamental ways that will solve the puzzle4 ,can you find it8 #cja4--,,the ,,eight ,,stars4 ,illustration3 ------------------------ _/_/_/ _/_/_/ ------------------------ _/_/_/ _/_/_/ 99 ------------------------ _/_/_/ _/_/_/ ------------------------ _/_/_/_/_/_/ ------------------------ _/_/_/_/_/_/ ------------------------ _/_/_/ _/_/_/ ------------------------ _/_/_/ _/_/_/ ------------------------ _/_/_/ _/_/_/ ------------------------ ,the puzzle in this case is to place eight stars in the diagram so that no star shall be in line with another star horizontally1 vertically1 or diagonally4 ,one star is already placed1 and that must not be moved1 so there are only seven for the reader now to place4 ,but you must not place a star on any one of the shaded squares4 ,there is only one way of solving this little puzzle4 #cjb4--;,a ,,problem ,,in ,,mosaics4 ,the art of producing pictures or designs by means of joining together pieces of hard substances1 either naturally or artificially coloured1 is of very great antiquity4 ,it was certainly known in the time of the ,pharaohs1 and we find a reference in the ,book of ,esther to a pavement of red1 and blue1 and white1 and black marble4 ,some of this ancient work that has come down to us1 especially some of the ,roman mosaics1 would seem to show clearly1 even where design is not at first evident1 that much thought was bestowed upon apparently disorderly arrangements4 ,where1 for example1 the work has been produced with a very limited number of colours1 there are evidences of great ingenuity in preventing the same tints coming in close proximity4 ,lady readers who are familiar with the construction of patchwork quilts will know how desirable it is sometimes1 when they are limited in the choice of material1 to prevent pieces of the same stuff coming too near together4 ,now1 this puzzle will apply equally to patchwork quilts or tesselated pavements4 ,it will be seen from the diagram how a square piece of flooring may be paved with sixty-two square tiles of the eight colours violet1 red1 yellow1 green1 orange1 purple1 white1 and blue 7indicated by the initial letters71 so that no tile is in line with a similarly coloured tile1 vertically1 horizontally1 or diagonally4 ,sixty-four such tiles could not possibly be placed under these conditions1 but the two shaded squares happen to be occupied by iron ventilators4 ,illustration3 ------------------------ ,v ,r ,y ,g ,o ,p ,w ,b ------------------------ ,w ,b ,o ,p ,y ,g ,v ,r ------9999------9999------ ,g ,p ,h ,w ,h ,v ,b ,h ,r ,h ,y ,o ------9999------9999------ ,r ,y ,b ,o ,g ,v ,p ,w ------------------------ ,b ,g ,r ,y ,p ,w ,o ,v ------------------------ ,o ,v ,p ,w ,r ,y ,b ,g ------------------------ ,p ,w ,g ,b ,v ,o ,r ,y ------------------------ _/_/_/ ,o ,v ,r ,w ,b ,g _/_/_/ ------------------------ ,the puzzle is this4 ,these two ventilators have to be removed to the positions indicated by the darkly bordered tiles1 and two tiles placed in those bottom corner squares4 ,can you readjust the thirty-two tiles so that no two of the same colour shall still be in line8 #cjc4--,,under ,,the ,,veil4 ,illustration3 ------------------------ ,v ,e ,i ,l ------------------------ ,i ,l ,v ,e ------------------------ ,i ,v ,l ,e ------------------------ ,l ,e ,i ,v ------------------------ ,v ,i ,e ,l ------------------------ ,e ,l ,v ,i ------------------------ ,e ,v ,l ,i ------------------------ ,l ,i ,e ,v ------------------------ ,if the reader will examine the above diagram1 he will see that ,i have so placed eight ,v's1 eight ,e's1 eight ,i's1 and eight ,l's in the diagram that no letter is in line with a similar one horizontally1 vertically1 or diagonally4 ,thus1 no ,v is in line with another ,v1 no ,e with another ,e1 and so on4 ,there are a great many different ways of arranging the letters under this condition4 ,the puzzle is to find an arrangement that produces the greatest possible number of four-letter words1 reading upwards and downwards1 backwards and forwards1 or diagonally4 ,all repetitions count as different words1 and the five variations that may be used are3 ,,veil1 ,,vile1 ,,levi1 ,,live1 and ,,evil4 ,this will be made perfectly clear when ,i say that the above arrangement scores eight1 because the top and bottom row both give ,,veil2 the second and seventh columns both give ,,veil2 and the two diagonals1 starting from the ,l in the #eth row and ,e in the #hth row1 both give ,,live and ,,evil4 ,there are therefore eight different readings of the words in all4 ,this difficult word puzzle is given as an example of the use of chessboard analysis in solving such things4 ,only a person who is familiar with the ,eight ,queens problem could hope to solve it4 #cjd4--;,,bachet',s ,,square4 ,one of the oldest card puzzles is by ,claude ,caspar ,bachet de ,mziriac1 first published1 ,i believe1 in the #afbd edition of his work4 ,rearrange the sixteen court cards 7including the aces7 in a square so that in no row of four cards1 horizontal1 vertical1 or diagonal1 shall be found two cards of the same suit or the same value4 ,this in itself is easy enough1 but a point of the puzzle is to find in how many different ways this may be done4 ,the eminent ,french mathematician ,a4 ,labosne1 in his modern edition of ,bachet1 gives the answer incorrectly4 ,and yet the puzzle is really quite easy4 ,any arrangement produces seven more by turning the square round and reflecting it in a mirror4 ,these are counted as different by ,bachet4 ,note row of four cards1 so that the only diagonals we have here to consider are the two long ones4 #cje4--,,the ,,thirty-,,six ,,letter-,,blocks4 ,illustration ,the illustration represents a box containing thirty-six letter-blocks4 ,the puzzle is to rearrange these blocks so that no ,a shall be in a line vertically1 horizontally1 or diagonally with another ,a1 no ,b with another ,b1 no ,c with another ,c1 and so on4 ,you will find it impossible to get all the letters into the box under these conditions1 but the point is to place as many as possible4 ,of course no letters other than those shown may be used4 #cjf4--,,the ,,crowded ,,chessboard4 ,illustration ,the puzzle is to rearrange the fifty-one pieces on the chessboard so that no queen shall attack another queen1 no rook attack another rook1 no bishop attack another bishop1 and no knight attack another knight4 ,no notice is to be taken of the intervention of pieces of another type from that under consideration--that is1 two queens will be considered to attack one another although there may be1 say1 a rook1 a bishop1 and a knight between them4 ,and so with the rooks and bishops4 ,it is not difficult to dispose of each type of piece separately2 the difficulty comes in when you have to find room for all the arrangements on the board simultaneously4 #cjg4--,,the ,,coloured ,,counters4 ,illustration ,the diagram represents twenty-five coloured counters1 ,red1 ,blue1 ,yellow1 ,orange1 and ,green 7indicated by their initials71 and there are five of each colour1 numbered #a1 #b1 #c1 #d1 and #e4 ,the problem is so to place them in a square that neither colour nor number shall be found repeated in any one of the five rows1 five columns1 and two diagonals4 ,can you so rearrange them8 #cjh4--,,the ,,gentle ,,art ,,of ,,stamp-,,licking4 ,the ,insurance ,act is a most prolific source of entertaining puzzles1 particularly entertaining if you happen to be among the exempt4 ,one's initiation into the gentle art of stamp-licking suggests the following little poser3 ,if you have a card divided into sixteen spaces 7#d #d71 and are provided with plenty of stamps of the values #a;d41 #b;d41 #c;d41 #d;d41 and #e;d41 what is the greatest value that you can stick on the card if the ,chancellor of the ,exchequer forbids you to place any stamp in a straight line 7that is1 horizontally1 vertically1 or diagonally7 with another stamp of similar value8 ,of course1 only one stamp can be affixed in a space4 ,the reader will probably find1 when he sees the solution1 that1 like the stamps themselves1 he is licked ,he will most likely be twopence short of the maximum4 ,a friend asked the ,post ,office how it was to be done2 but they sent him to the ,customs and ,excise officer1 who sent him to the ,insurance ,commissioners1 who sent him to an approved society1 who profanely sent him--but no matter4 #cji4--,,the ,,forty-,,nine ,,counters4 ,illustration ,can you rearrange the above forty-nine counters in a square so that no letter1 and also no number1 shall be in line with a similar one1 vertically1 horizontally1 or diagonally8 ,here ,i1 of course1 mean in the lines parallel with the diagonals1 in the chessboard sense4 #caj4--,,the ,,three ,,sheep4 ,illustration ,a farmer had three sheep and an arrangement of sixteen pens1 divided off by hurdles in the manner indicated in the illustration4 ,in how many different ways could he place those sheep1 each in a separate pen1 so that every pen should be either occupied or in line 7horizontally1 vertically1 or diagonally7 with at least one sheep8 ,i have given one arrangement that fulfils the conditions4 ,how many others can you find8 ,mere reversals and reflections must not be counted as different4 ,the reader may regard the sheep as queens4 ,the problem is then to place the three queens so that every square shall be either occupied or attacked by at least one queen--in the maximum number of different ways4 #caa4--,,the ,,five ,,dogs ,,puzzle4 ,in #ahfc1 ,c4,f4 de ,jaenisch first discussed the ,five ,queens ,puzzle--to place five queens on the chessboard so that every square shall be attacked or occupied--which was propounded by his friend1 a ,mr4 de ,r4 ,jaenisch showed that if no queen may attack another there are ninety-one different ways of placing the five queens1 reversals and reflections not counting as different4 ,if the queens may attack one another1 ,i have recorded hundreds of ways1 but it is not practicable to enumerate them exactly4 ,illustration ,the illustration is supposed to represent an arrangement of sixty-four kennels4 ,it will be seen that five kennels each contain a dog1 and on further examination it will be seen that every one of the sixty-four kennels is in a straight line with at least one dog--either horizontally1 vertically1 or diagonally4 ,take any kennel you like1 and you will find that you can draw a straight line to a dog in one or other of the three ways mentioned4 ,the puzzle is to replace the five dogs and discover in just how many different ways they may be placed in five kennels in a straight row1 so that every kennel shall always be in line with at least one dog4 ,reversals and reflections are here counted as different4 #cab4--,,the ,,five ,,crescents ,,of ,,byzantium4 ,when ,philip of ,macedon1 the father of ,alexander the ,great1 found himself confronted with great difficulties in the siege of ,byzantium1 he set his men to undermine the walls4 ,his desires1 however1 miscarried1 for no sooner had the operations been begun than a crescent moon suddenly appeared in the heavens and discovered his plans to his adversaries4 ,the ,byzantines were naturally elated1 and in order to show their gratitude they erected a statue to ,diana1 and the crescent became thenceforward a symbol of the state4 ,in the temple that contained the statue was a square pavement composed of sixty-four large and costly tiles4 ,these were all plain1 with the exception of five1 which bore the symbol of the crescent4 ,these five were for occult reasons so placed that every tile should be watched over by 7that is1 in a straight line1 vertically1 horizontally1 or diagonally with7 at least one of the crescents4 ,the arrangement adopted by the ,byzantine architect was as follows3-- ,illustration ,now1 to cover up one of these five crescents was a capital offence1 the death being something very painful and lingering4 ,but on a certain occasion of festivity it was necessary to lay down on this pavement a square carpet of the largest dimensions possible1 and ,i have shown in the illustration by dark shading the largest dimensions that would be available4 ,the puzzle is to show how the architect1 if he had foreseen this question of the carpet1 might have so arranged his five crescent tiles in accordance with the required conditions1 and yet have allowed for the largest possible square carpet to be laid down without any one of the five crescent tiles being covered1 or any portion of them4 #cac4--,,queens ,,and ,,bishop ,,puzzle4 ,it will be seen that every square of the board is either occupied or attacked4 ,the puzzle is to substitute a bishop for the rook on the same square1 and then place the four queens on other squares so that every square shall again be either occupied or attacked4 ,illustration #cad4--,,the ,,southern ,,cross4 ,illustration ,in the above illustration we have five ,planets and eighty-one ,fixed ,stars1 five of the latter being hidden by the ,planets4 ,it will be found that every ,star1 with the exception of the ten that have a black spot in their centres1 is in a straight line1 vertically1 horizontally1 or diagonally1 with at least one of the ,planets4 ,the puzzle is so to rearrange the ,planets that all the ,stars shall be in line with one or more of them4 ,in rearranging the ,planets1 each of the five may be moved once in a straight line1 in either of the three directions mentioned4 ,they will1 of course1 obscure five other ,stars in place of those at present covered4 #cae4--,,the ,,hat-,,peg ,,puzzle4 ,here is a five-queen puzzle that ,i gave in a fanciful dress in #ahig4 ,as the queens were there represented as hats on sixty-four pegs1 ,i will keep to the title1 ,the ,hat-,peg ,puzzle4 ,it will be seen that every square is occupied or attacked4 ,the puzzle is to remove one queen to a different square so that still every square is occupied or attacked1 then move a second queen under a similar condition1 then a third queen1 and finally a fourth queen4 ,after the fourth move every square must be attacked or occupied1 but no queen must then attack another4 ,of course1 the moves need not be queen moves2 you can move a queen to any part of the board4 ,illustration #caf4--,,the ,,amazons4 ,illustration ,this puzzle is based on one by ,captain ,turton4 ,remove three of the queens to other squares so that there shall be eleven squares on the board that are not attacked4 ,the removal of the three queens need not be by queen moves4 ,you may take them up and place them anywhere4 ,there is only one solution4 #cag4--;,a ,,puzzle ,,with ,,pawns4 ,place two pawns in the middle of the chessboard1 one at ,q #d and the other at ,k #e4 ,now1 place the remaining fourteen pawns 7sixteen in all7 so that no three shall be in a straight line in any possible direction4 ,note that ,i purposely do not say queens1 because by the words any possible direction ,i go beyond attacks on diagonals4 ,the pawns must be regarded as mere points in space--at the centres of the squares4 ,see dotted lines in the case of ,no4 #cjj1 ,the ,eight ,queens4 #cah4--,,lion-,,hunting4 ,illustration ,my friend ,captain ,potham ,hall1 the renowned hunter of big game1 says there is nothing more exhilarating than a brush with a herd--a pack--a team--a flock--a swarm 7it has taken me a full quarter of an hour to recall the right word1 but ,i have it at last7--a pride of lions4 ,why a number of lions are called a pride1 a number of whales a school1 and a number of foxes a skulk are mysteries of philology into which ,i will not enter4 ,well1 the captain says that if a spirited lion crosses your path in the desert it becomes lively1 for the lion has generally been looking for the man just as much as the man has sought the king of the forest4 ,and yet when they meet they always quarrel and fight it out4 ,a little contemplation of this unfortunate and long-standing feud between two estimable families has led me to figure out a few calculations as to the probability of the man and the lion crossing one another's path in the jungle4 ,in all these cases one has to start on certain more or less arbitrary assumptions4 ,that is why in the above illustration ,i have thought it necessary to represent the paths in the desert with such rigid regularity4 ,though the captain assures me that the tracks of the lions usually run much in this way1 ,i have doubts4 ,the puzzle is simply to find out in how many different ways the man and the lion may be placed on two different spots that are not on the same path4 ,by paths it must be understood that ,i only refer to the ruled lines4 ,thus1 with the exception of the four corner spots1 each combatant is always on two paths and no more4 ,it will be seen that there is a lot of scope for evading one another in the desert1 which is just what one has always understood4 #cai4--,,the ,,knight-,,guards4 ,illustration ,the knight is the irresponsible low comedian of the chessboard4 ,he is a very uncertain1 sneaking1 and demoralizing rascal1 says an ,american writer4 ,he can only move two squares1 but makes up in the quality of his locomotion for its quantity1 for he can spring one square sideways and one forward simultaneously1 like a cat2 can stand on one leg in the middle of the board and jump to any one of eight squares he chooses2 can get on one side of a fence and blackguard three or four men on the other2 has an objectionable way of inserting himself in safe places where he can scare the king and compel him to move1 and then gobble a queen4 ,for pure cussedness the knight has no equal1 and when you chase him out of one hole he skips into another4 ,attempts have been made over and over again to obtain a short1 simple1 and exact definition of the move of the knight--without success4 ,it really consists in moving one square like a rook1 and then another square like a bishop--the two operations being done in one leap1 so that it does not matter whether the first square passed over is occupied by another piece or not4 ,it is1 in fact1 the only leaping move in chess4 ,but difficult as it is to define1 a child can learn it by inspection in a few minutes4 ,i have shown in the diagram how twelve knights 7the fewest possible that will perform the feat7 may be placed on the chessboard so that every square is either occupied or attacked by a knight4 ,examine every square in turn1 and you will find that this is so4 ,now1 the puzzle in this case is to discover what is the smallest possible number of knights that is required in order that every square shall be either occupied or attacked1 and every knight protected by another knight4 ,and how would you arrange them8 ,it will be found that of the twelve shown in the diagram only four are thus protected by being a knight's move from another knight4 ,,the ,,guarded ,,chessboard4 ,on an ordinary chessboard1 #h by #h1 every square can be guarded--that is1 either occupied or attacked--by #e queens1 the fewest possible4 ,there are exactly #ia fundamentally different arrangements in which no queen attacks another queen4 ,if every queen must attack 7or be protected by7 another queen1 there are at fewest #da arrangements1 and ,i have recorded some #aej ways in which some of the queens are attacked and some not1 but this last case is very difficult to enumerate exactly4 ,on an ordinary chessboard every square can be guarded by #h rooks 7the fewest possible7 in #dj1cbj ways1 if no rook may attack another rook1 but it is not known how many of these are fundamentally different4 7,see solution to ,no4 #bie1 ,the ,eight ,rooks47 ,i have not enumerated the ways in which every rook shall be protected by another rook4 ,on an ordinary chessboard every square can be guarded by #h bishops 7the fewest possible71 if no bishop may attack another bishop4 ,ten bishops are necessary if every bishop is to be protected4 7,see ,nos4 #big and #bih1 ,bishops unguarded and ,bishops guarded47 ,on an ordinary chessboard every square can be guarded by #ab knights if all but #d are unprotected4 ,but if every knight must be protected1 #ad are necessary4 7,see ,no4 #cai1 ,the ,knight-,guards47 ,dealing with the queen on n boards generally1 where n is less than #h1 the following results will be of interest3-- #a queen guards #b board in #a fundamental way4 #a queen guards #c board in #a fundamental way4 #b queens guard #d board in #c fundamental ways 7protected74 #c queens guard #d board in #b fundamental ways 7not protected74 #c queens guard #e board in #cg fundamental ways 7protected74 #c queens guard #e board in #b fundamental ways 7not protected74 #c queens guard #f board in #a fundamental way 7protected74 #d queens guard #f board in #ag fundamental ways 7not protected74 #d queens guard #g board in #e fundamental ways 7protected74 #d queens guard #g board in #a fundamental way 7not protected74 ,,non-,,attacking ,,chessboard ,,arrangements4 ,we know that n queens may always be placed on a square board of n squares 7if n be greater than #c7 without any queen attacking another queen4 ,but no general formula for enumerating the number of different ways in which it may be done has yet been discovered2 probably it is undiscoverable4 ,the known results are as follows3-- ,where n #d there is #a fundamental solution and #b in all4 ,where n #e there are #b fundamental solutions and #aj in all4 ,where n #f there is #a fundamental solution and #d in all4 ,where n #g there are #f fundamental solutions and #dj in all4 ,where n #h there are #ab fundamental solutions and #ib in all4 ,where n #i there are #df fundamental solutions4 ,where n #aj there are #ib fundamental solutions4 ,where n #aa there are #cda fundamental solutions4 ,obviously n rooks may be placed without attack on an n board in n6 ways1 but how many of these are fundamentally different ,i have only worked out in the four cases where n equals #b1 #c1 #d1 and #e4 ,the answers here are respectively #a1 #b1 #g1 and #bc4 7,see ,no4 #bif1 ,the ,four ,lions47 ,we can place #bn-#b bishops on an n board in #bn ways4 7,see ,no4 #bii1 ,bishops in ,convocation47 ,for boards containing #b1 #c1 #d1 #e1 #f1 #g1 #h squares1 on a side there are respectively #a1 #b1 #c1 #f1 #aj1 #bj1 #cf fundamentally different arrangements4 ,where n is odd there are #b7n-#a7 such arrangements1 each giving #d by reversals and reflections1 and #bn-#c - #b7n-#c7 giving #h4 ,where n is even there are #b7n-#b71 each giving #d by reversals and reflections1 and #bn-#c - #b7n-#d71 each giving #h4 ,we can place 7n#a7 knights on an n board without attack1 when n is odd1 in #a fundamental way2 and n knights on an n board1 when n is even1 in #a fundamental way4 ,in the first case we place all the knights on the same colour as the central square2 in the second case we place them all on black1 or all on white1 squares4 ,,the ,,two ,,pieces ,,problem4 ,on a board of n squares1 two queens1 two rooks1 two bishops1 or two knights can always be placed1 irrespective of attack or not1 in 7n#d - n7 ways4 ,the following formul will show in how many of these ways the two pieces may be placed with attack and without3-- ,with ,attack4 ,without ,attack4 #b ,queens #en - #fn n #cn#d - #ajn #in - #bn ------------------- ------------------------------ #c #f #b ,rooks n - n n#d - #bn n ---------------------- #b #b ,bishops #dn - #fn #bn #cn#d - #dn #cn - #bn -------------------- ----------------------------- #f #f #b ,knights #dn - #abn #h n#d - #in #bdn -------------------- #b 7,see ,no4 #cah1 ,lion ,hunting47 ,,dynamical ,,chess ,,puzzles4 ,push on--keep moving4 ,,thos4 ,,morton3 ,cure for the ,heartache4 #cbj4--,,the ,,rook',s ,,tour4 ,illustration3 ------------------------ ------------------------ ------------------------ ------------------------ ,r ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ,the puzzle is to move the single rook over the whole board1 so that it shall visit every square of the board once1 and only once1 and end its tour on the square from which it starts4 ,you have to do this in as few moves as possible1 and unless you are very careful you will take just one move too many4 ,of course1 a square is regarded equally as visited whether you merely pass over it or make it a stopping-place1 and we will not quibble over the point whether the original square is actually visited twice4 ,we will assume that it is not4 #cba4--,,the ,,rook',s ,,journey4 ,this puzzle ,i call ,the ,rook's ,journey1 because the word tour 7derived from a turner's wheel7 implies that we return to the point from which we set out1 and we do not do this in the present case4 ,we should not be satisfied with a personally conducted holiday tour that ended by leaving us1 say1 in the middle of the ,sahara4 ,the rook here makes twenty-one moves1 in the course of which journey it visits every square of the board once and only once1 stopping at the square marked #aj at the end of its tenth move1 and ending at the square marked #ba4 ,two consecutive moves cannot be made in the same direction--that is to say1 you must make a turn after every move4 ,illustration3 ------------------------ ,r ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ ------------------------ #ba #aj ------------------------ #cbb4--,,the ,,languishing ,,maiden4 ,illustration3 ------------------------------------- ,kt -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- ,m -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- ,a wicked baron in the good old days imprisoned an innocent maiden in one of the deepest dungeons beneath the castle moat4 ,it will be seen from our illustration that there were sixty-three cells in the dungeon1 all connected by open doors1 and the maiden was chained in the cell in which she is shown4 ,now1 a valiant knight1 who loved the damsel1 succeeded in rescuing her from the enemy4 ,having gained an entrance to the dungeon at the point where he is seen1 he succeeded in reaching the maiden after entering every cell once and only once4 ,take your pencil and try to trace out such a route4 ,when you have succeeded1 then try to discover a route in twenty-two straight paths through the cells4 ,it can be done in this number without entering any cell a second time4 #cbc4--;,a ,,dungeon ,,puzzle4 ,illustration3 ---------------------------------------- 4444444444444 4444444 4444444444444 4 4 4 4 4 4 --4---- ----4----4----4----4---- ----4-- 4 4 4 4 4 4 4444444 4444444 4444444 4444444 4 4 -- ----4---- ---- ---- ---- ----4---- -- 4 4 4444444 4444444 4444444 4444444 4 4 4 4 4 4 --4---- ----4----4----4----4---- ----4-- 4 4 4 4 4 4 4444444444444 4444444 4 4444444 4 4 -- ---- ---- ---- ---- ----4----4---- -- 4 4 4444444444444 4444444 4 4444444 4 4 4 4 4 4 --4---- ----4----4----4----4---- ----4-- 4 4 4 4 4 4 4444444 4444444 4444444 4444444 4 4 -- ----4---- ---- ---- ---- ----4---- -- 4 4 4444444 4444444 4444444 4444444 4 4 4 4 4 4 --4---- ----4----4----4----4---- ----4-- 4 4 4 4 4 4 4444444444444 4 ,p 4444444444444 -- ---- ---- ---- ---- ---- ---- ---- -- ,a ,french prisoner1 for his sins 7or other people's71 was confined in an underground dungeon containing sixty-four cells1 all communicating with open doorways1 as shown in our illustration4 ,in order to reduce the tedium of his restricted life1 he set himself various puzzles1 and this is one of them4 ,starting from the cell in which he is shown1 how could he visit every cell once1 and only once1 and make as many turnings as possible8 ,his first attempt is shown by the dotted track4 ,it will be found that there are as many as fifty-five straight lines in his path1 but after many attempts he improved upon this4 ,can you get more than fifty-five8 ,you may end your path in any cell you like4 ,try the puzzle with a pencil on chessboard diagrams1 or you may regard them as rooks' moves on a board4 #cbd4--,,the ,,lion ,,and ,,the ,,man4 ,in a public place in ,rome there once stood a prison divided into sixty-four cells1 all open to the sky and all communicating with one another1 as shown in the illustration4 ,the sports that here took place were watched from a high tower4 ,the favourite game was to place a ,christian in one corner cell and a lion in the diagonally opposite corner and then leave them with all the inner doors open4 ,the consequent effect was sometimes most laughable4 ,on one occasion the man was given a sword4 ,he was no coward1 and was as anxious to find the lion as the lion undoubtedly was to find him4 ,illustration3 ---------------------------------------- ,l -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- -- ---- ---- ---- ---- ---- ---- ---- -- ,c -- ---- ---- ---- ---- ---- ---- ---- -- ,the man visited every cell once and only once in the fewest possible straight lines until he reached the lion's cell4 ,the lion1 curiously enough1 also visited every cell once and only once in the fewest possible straight lines until he finally reached the man's cell4 ,they started together and went at the same speed2 yet1 although they occasionally got glimpses of one another1 they never once met4 ,the puzzle is to show the route that each happened to take4 #cbe4--;,,an ,,episcopal ,,visitation4 ,the white squares on the chessboard represent the parishes of a diocese4 ,place the bishop on any square you like1 and so contrive that 7using the ordinary bishop's move of chess7 he shall visit every one of his parishes in the fewest possible moves4 ,of course1 all the parishes passed through on any move are regarded as visited4 ,you can visit any squares more than once1 but you are not allowed to move twice between the same two adjoining squares4 ,what are the fewest possible moves8 ,the bishop need not end his visitation at the parish from which he first set out4 #cbf4--;,a ,,new ,,counter ,,puzzle4 ,here is a new puzzle with moving counters1 or coins1 that at first glance looks as if it must be absurdly simple4 ,but it will be found quite a little perplexity4 ,i give it in this place for a reason that ,i will explain when we come to the next puzzle4 ,copy the simple diagram1 enlarged1 on a sheet of paper2 then place two white counters on the points #a and #b1 and two red counters on #i and #aj1 ,the puzzle is to make the red and white change places4 ,you may move the counters one at a time in any order you like1 along the lines from point to point1 with the only restriction that a red and a white counter may never stand at once on the same straight line4 ,thus the first move can only be from #a or #b to #c1 or from #i or #aj to #g4 ,illustration3 #d #h _/ _/ #b #f #aj _/ _/ #c #g _/ _/ #a #e #i #cbg4--;,a ,,new ,,bishop',s ,,puzzle4 ,illustration3 ------------ b b b b ------------ ------------ ------------ ,b ,b ,b ,b ------------ ,this is quite a fascinating little puzzle4 ,place eight bishops 7four black and four white7 on the reduced chessboard1 as shown in the illustration4 ,the problem is to make the black bishops change places with the white ones1 no bishop ever attacking another of the opposite colour4 ,they must move alternately--first a white1 then a black1 then a white1 and so on4 ,when you have succeeded in doing it at all1 try to find the fewest possible moves4 ,if you leave out the bishops standing on black squares1 and only play on the white squares1 you will discover my last puzzle turned on its side4 #cbh4--,,the ,,queen',s ,,tour4 ,the puzzle of making a complete tour of the chessboard with the queen in the fewest possible moves 7in which squares may be visited more than once7 was first given by the late ,sam ,loyd in his ,chess ,strategy4 ,but the solution shown below is the one he gave in ,american ,chess-,nuts in #ahfh4 ,i have recorded at least six different solutions in the minimum number of moves--fourteen--but this one is the best of all1 for reasons ,i will explain4 ,illustration3 ------------------------ 44444444444444444444444444444 4 4 -4--------------------4- 4 4 4 44444444444444444444444444 4 4 4 -4----4---------------44- 4 4 4 4 4 44444444444444444 4 4 4 4 4 4 4 4 -4----4---4------4---4---4- 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 44 4 44 4 44 4 -4--4-4---4---4---4--4-4--4- 4 4 4 4 44 4 4 4 4 4 4 4 4 44 4 4 4 4 4 44 4 444 4 4 -4--4----4---44--4----4--4- 4 4 44 44 44 4 4 4 4 4 44 4 4 4 4 4 4 4 444 4 4 4 4 -4--4-4---4--4---4---4-4--4- 4 44 4 44 4 44 4 4 4 4 4 4 4 4 4 4 44 4 4 4 4 44 4 -4-4-4-4--4------4---4-4-4-4- 44 4 4 4 4 44 44 4 44 4444444444444 4 4 4 ------------------------ ,if you will look at the lettered square you will understand that there are only ten really differently placed squares on a chessboard--those enclosed by a dark line--all the others are mere reversals or reflections4 ,for example1 every ,a is a corner square1 and every ,j a central square4 ,consequently1 as the solution shown has a turning-point at the enclosed ,d square1 we can obtain a solution starting from and ending at any square marked ,d--by just turning the board about4 ,now1 this scheme will give you a tour starting from any ,a1 ,b1 ,c1 ,d1 ,e1 ,f1 or ,h1 while no other route that ,i know can be adapted to more than five different starting-points4 ,there is no ,queen's ,tour in fourteen moves 7remember a tour must be re-entrant7 that may start from a ,g1 ,i1 or ,j4 ,but we can have a non-re-entrant path over the whole board in fourteen moves1 starting from any given square4 ,hence the following puzzle3-- ,illustration3 ------------99------------ ,a ,b ,c ,g ,g ,c ,b ,a 9999---------99------------ ,b ,d ,e ,h ,h ,e ,d ,b ---9999------99------------ ,c ,e ,f ,i ,i ,f ,e ,c ------9999---99------------ ,g ,h ,i ,j ,j ,i ,h ,g ---------9999------------ ,g ,h ,i ,j ,j ,i ,h ,g ------------------------ ,c ,e ,f ,i ,i ,f ,e ,c ------------------------ ,b ,d ,e ,h ,h ,e ,d ,b ------------------------ ,a ,b ,c ,g ,g ,c ,b ,a ------------------------ ,start from the ,j in the enclosed part of the lettered diagram and visit every square of the board in fourteen moves1 ending wherever you like4 #cbi4--,,the ,,star ,,puzzle4 ,illustration3 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ 99 99 99 99 99 99 99 99 ------------------------ ,put the point of your pencil on one of the white stars and 7without ever lifting your pencil from the paper7 strike out all the stars in fourteen continuous straight strokes1 ending at the second white star4 ,your straight strokes may be in any direction you like1 only every turning must be made on a star4 ,there is no objection to striking out any star more than once4 ,in this case1 where both your starting and ending squares are fixed inconveniently1 you cannot obtain a solution by breaking a ,queen's ,tour1 or in any other way by queen moves alone4 ,but you are allowed to use oblique straight lines--such as from the upper white star direct to a corner star4 #ccj4--,,the ,,yacht ,,race4 ,now then1 ye land-lubbers1 hoist your baby-jib-topsails1 break out your spinnakers1 ease off your balloon sheets1 and get your head-sails set6 ,our race consists in starting from the point at which the yacht is lying in the illustration and touching every one of the sixty-four buoys in fourteen straight courses1 returning in the final tack to the buoy from which we start4 ,the seventh course must finish at the buoy from which a flag is flying4 ,this puzzle will call for a lot of skilful seamanship on account of the sharp angles at which it will occasionally be necessary to tack4 ,the point of a lead pencil and a good nautical eye are all the outfit that we require4 ,illustration ,this is difficult1 because of the condition as to the flag-buoy1 and because it is a re-entrant tour4 ,but again we are allowed those oblique lines4 #cca4--,,the ,,scientific ,,skater4 ,illustration ,it will be seen that this skater has marked on the ice sixty-four points or stars1 and he proposes to start from his present position near the corner and enter every one of the points in fourteen straight lines4 ,how will he do it8 ,of course there is no objection to his passing over any point more than once1 but his last straight stroke must bring him back to the position from which he started4 ,it is merely a matter of taking your pencil and starting from the spot on which the skater's foot is at present resting1 and striking out all the stars in fourteen continuous straight lines1 returning to the point from which you set out4 #ccb4--,,the ,,forty-,,nine ,,stars4 ,illustration ,the puzzle in this case is simply to take your pencil and1 starting from one black star1 strike out all the stars in twelve straight strokes1 ending at the other black star4 ,it will be seen that the attempt shown in the illustration requires fifteen strokes4 ,can you do it in twelve8 ,every turning must be made on a star1 and the lines must be parallel to the sides and diagonals of the square1 as shown4 ,in this case we are dealing with a chessboard of reduced dimensions1 but only queen moves 7without going outside the boundary as in the last case7 are required4 #ccc4--,,the ,,queen',s ,,journey4 ,illustration ,place the queen on her own square1 as shown in the illustration1 and then try to discover the greatest distance that she can travel over the board in five queen's moves without passing over any square a second time4 ,mark the queen's path on the board1 and note carefully also that she must never cross her own track4 ,it seems simple enough1 but the reader may find that he has tripped4 #ccd4--,,st4 ,,george ,,and ,,the ,,dragon4 ,illustration ,here is a little puzzle on a reduced chessboard of forty-nine squares4 ,st4 ,george wishes to kill the dragon4 ,killing dragons was a well-known pastime of his1 and1 being a knight1 it was only natural that he should desire to perform the feat in a series of knight's moves4 ,can you show how1 starting from that central square1 he may visit once1 and only once1 every square of the board in a chain of chess knight's moves1 and end by capturing the dragon on his last move8 ,of course a variety of different ways are open to him1 so try to discover a route that forms some pretty design when you have marked each successive leap by a straight line from square to square4 #cce4--;,,farmer ,,lawrence',s ,,cornfields4 ,one of the most beautiful districts within easy distance of ,london for a summer ramble is that part of ,buckinghamshire known as the ,valley of the ,chess--at least1 it was a few years ago1 before it was discovered by the speculative builder4 ,at the beginning of the present century there lived1 not far from ,latimers1 a worthy but eccentric farmer named ,lawrence4 ,one of his queer notions was that every person who lived near the banks of the river ,chess ought to be in some way acquainted with the noble game of the same name1 and in order to impress this fact on his men and his neighbours he adopted at times strange terminology4 ,for example1 when one of his ewes presented him with a lamb1 he would say that it had queened a pawn2 when he put up a new barn against the highway1 he called it castling on the king's side2 and when he sent a man with a gun to keep his neighbour's birds off his fields1 he spoke of it as attacking his opponent's rooks4 ,everybody in the neighbourhood used to be amused at ,farmer ,lawrence's little jokes1 and one boy 7the wag of the village7 who got his ears pulled by the old gentleman for stealing his chestnuts went so far as to call him a silly old chess-protector6 ,one year he had a large square field divided into forty-nine square plots1 as shown in the illustration4 ,the white squares were sown with wheat and the black squares with barley4 ,when the harvest time came round he gave orders that his men were first to cut the corn in the patch marked #a1 and that each successive cutting should be exactly a knight's move from the last one1 the thirteenth cutting being in the patch marked #ac1 the twenty-fifth in the patch marked #be1 the thirty-seventh in the one marked #cg1 and the last1 or forty-ninth cutting1 in the patch marked #di4 ,this was too much for poor ,hodge1 and each day ,farmer ,lawrence had to go down to the field and show which piece had to be operated upon4 ,but the problem will perhaps present no difficulty to my readers4 ,illustration #ccf4--,,the ,,greyhound ,,puzzle4 ,in this puzzle the twenty kennels do not communicate with one another by doors1 but are divided off by a low wall4 ,the solitary occupant is the greyhound which lives in the kennel in the top left-hand corner4 ,when he is allowed his liberty he has to obtain it by visiting every kennel once and only once in a series of knight's moves1 ending at the bottom right-hand corner1 which is open to the world4 ,the lines in the above diagram show one solution4 ,the puzzle is to discover in how many different ways the greyhound may thus make his exit from his corner kennel4 ,illustration #ccg4--,,the ,,four ,,kangaroos4 ,illustration ,in introducing a little ,commonwealth problem1 ,i must first explain that the diagram represents the sixty-four fields1 all properly fenced off from one another1 of an ,australian settlement1 though ,i need hardly say that our kith and kin down under always do set out their land in this methodical and exact manner4 ,it will be seen that in every one of the four corners is a kangaroo4 ,why kangaroos have a marked preference for corner plots has never been satisfactorily explained1 and it would be out of place to discuss the point here4 ,i should also add that kangaroos1 as is well known1 always leap in what we call knight's moves4 ,in fact1 chess players would probably have adopted the better term kangaroo's move had not chess been invented before kangaroos4 ,the puzzle is simply this4 ,one morning each kangaroo went for his morning hop1 and in sixteen consecutive knight's leaps visited just fifteen different fields and jumped back to his corner4 ,no field was visited by more than one of the kangaroos4 ,the diagram shows how they arranged matters4 ,what you are asked to do is to show how they might have performed the feat without any kangaroo ever crossing the horizontal line in the middle of the square that divides the board into two equal parts4 #cch4--,,the ,,board ,,in ,,compartments4 ,illustration ,we cannot divide the ordinary chessboard into four equal square compartments1 and describe a complete tour1 or even path1 in each compartment4 ,but we may divide it into four compartments1 as in the illustration1 two containing each twenty squares1 and the other two each twelve squares1 and so obtain an interesting puzzle4 ,you are asked to describe a complete re-entrant tour on this board1 starting where you like1 but visiting every square in each successive compartment before passing into another one1 and making the final leap back to the square from which the knight set out4 ,it is not difficult1 but will be found very entertaining and not uninstructive4 ,whether a re-entrant tour or a complete knight's path is possible or not on a rectangular board of given dimensions depends not only on its dimensions1 but also on its shape4 ,a tour is obviously not possible on a board containing an odd number of cells1 such as #e by #e or #g by #g1 for this reason3 ,every successive leap of the knight must be from a white square to a black and a black to a white alternately4 ,but if there be an odd number of cells or squares there must be one more square of one colour than of the other1 therefore the path must begin from a square of the colour that is in excess1 and end on a similar colour1 and as a knight's move from one colour to a similar colour is impossible the path cannot be re-entrant4 ,but a perfect tour may be made on a rectangular board of any dimensions provided the number of squares be even1 and that the number of squares on one side be not less than #f and on the other not less than #e4 ,in other words1 the smallest rectangular board on which a re-entrant tour is possible is one that is #f by #e4 ,a complete knight's path 7not re-entrant7 over all the squares of a board is never possible if there be only two squares on one side2 nor is it possible on a square board of smaller dimensions than #e by #e4 ,so that on a board #d by #d we can neither describe a knight's tour nor a complete knight's path2 we must leave one square unvisited4 ,yet on a board #d by #c 7containing four squares fewer7 a complete path may be described in sixteen different ways4 ,it may interest the reader to discover all these4 ,every path that starts from and ends at different squares is here counted as a different solution1 and even reverse routes are called different4 #cci4--,,the ,,four ,,knights' ,,tours4 ,illustration ,i will repeat that if a chessboard be cut into four equal parts1 as indicated by the dark lines in the illustration1 it is not possible to perform a knight's tour1 either re-entrant or not1 on one of the parts4 ,the best re-entrant attempt is shown1 in which each knight has to trespass twice on other parts4 ,the puzzle is to cut the board differently into four parts1 each of the same size and shape1 so that a re-entrant knight's tour may be made on each part4 ,cuts along the dotted lines will not do1 as the four central squares of the board would be either detached or hanging on by a mere thread4 #cdj4--,,the ,,cubic ,,knight',s ,,tour4 ,some few years ago ,i happened to read somewhere that ,abnit ,vandermonde1 a clever mathematician1 who was born in #agcf and died in #agic1 had devoted a good deal of study to the question of knight's tours4 ,beyond what may be gathered from a few fragmentary references1 ,i am not aware of the exact nature or results of his investigations1 but one thing attracted my attention1 and that was the statement that he had proposed the question of a tour of the knight over the six surfaces of a cube1 each surface being a chessboard4 ,whether he obtained a solution or not ,i do not know1 but ,i have never seen one published4 ,so ,i at once set to work to master this interesting problem4 ,perhaps the reader may like to attempt it4 #cda4--,,the ,,four ,,frogs4 ,illustration ,in the illustration we have eight toadstools1 with white frogs on #a and #c and black frogs on #f and #h4 ,the puzzle is to move one frog at a time1 in any order1 along one of the straight lines from toadstool to toadstool1 until they have exchanged places1 the white frogs being left on #f and #h and the black ones on #a and #c4 ,if you use four counters on a simple diagram1 you will find this quite easy1 but it is a little more puzzling to do it in only seven plays1 any number of successive moves by one frog counting as one play4 ,of course1 more than one frog cannot be on a toadstool at the same time4 #cdb4--,,the ,,mandarin',s ,,puzzle4 ,the following puzzle has an added interest from the circumstance that a correct solution of it secured for a certain young ,chinaman the hand of his charming bride4 ,the wealthiest mandarin within a radius of a hundred miles of ,peking was ,hi-,chum-,chop1 and his beautiful daughter1 ,peeky-,bo1 had innumerable admirers4 ,one of her most ardent lovers was ,winky-,hi1 and when he asked the old mandarin for his consent to their marriage1 ,hi-,chum-,chop presented him with the following puzzle and promised his consent if the youth brought him the correct answer within a week4 ,winky-,hi1 following a habit which obtains among certain solvers to this day1 gave it to all his friends1 and when he had compared their solutions he handed in the best one as his own4 ,luckily it was quite right4 ,the mandarin thereupon fulfilled his promise4 ,the fatted pup was killed for the wedding feast1 and when ,hi-,chum-,chop passed ,winky-,hi the liver wing all present knew that it was a token of eternal goodwill1 in accordance with ,chinese custom from time immemorial4 ,the mandarin had a table divided into twenty-five squares1 as shown in the diagram4 ,on each of twenty-four of these squares was placed a numbered counter1 just as ,i have indicated4 ,the puzzle is to get the counters in numerical order by moving them one at a time in what we call knight's moves4 ,counter #a should be where #af is1 #b where #aa is1 #d where #ac now is1 and so on4 ,it will be seen that all the counters on shaded squares are in their proper positions4 ,of course1 two counters may never be on a square at the same time4 ,can you perform the feat in the fewest possible moves8 ,illustration ,in order to make the manner of moving perfectly clear ,i will point out that the first knight's move can only be made by #a or by #b or by #aj4 ,supposing #a moves1 then the next move must be by #bc1 #d1 #h1 or #ba4 ,as there is never more than one square vacant1 the order in which the counters move may be written out as follows3 #a--ba--ad--ah--bb1 etc4 ,a rough diagram should be made on a larger scale for practice1 and numbered counters or pieces of cardboard used4 #cdc4--;,,exercise ,,for ,,prisoners4 ,the following is the plan of the north wing of a certain gaol1 showing the sixteen cells all communicating by open doorways4 ,fifteen prisoners were numbered and arranged in the cells as shown4 ,they were allowed to change their cells as much as they liked1 but if two prisoners were ever in the same cell together there was a severe punishment promised them4 ,illustration ,now1 in order to reduce their growing obesity1 and to combine physical exercise with mental recreation1 the prisoners decided1 on the suggestion of one of their number who was interested in knight's tours1 to try to form themselves into a perfect knight's path without breaking the prison regulations1 and leaving the bottom right-hand corner cell vacant1 as originally4 ,the joke of the matter is that the arrangement at which they arrived was as follows3-- #h #c #ab #a #aa #ad #i #f #d #g #b #ac #ae #aj #e ,the warders failed to detect the important fact that the men could not possibly get into this position without two of them having been at some time in the same cell together4 ,make the attempt with counters on a ruled diagram1 and you will find that this is so4 ,otherwise the solution is correct enough1 each member being1 as required1 a knight's move from the preceding number1 and the original corner cell vacant4 ,the puzzle is to start with the men placed as in the illustration and show how it might have been done in the fewest moves1 while giving a complete rest to as many prisoners as possible4 ,as there is never more than one vacant cell for a man to enter1 it is only necessary to write down the numbers of the men in the order in which they move4 ,it is clear that very few men can be left throughout in their cells undisturbed1 but ,i will leave the solver to discover just how many1 as this is a very essential part of the puzzle4 #cdd4--,,the ,,kennel ,,puzzle4 ,illustration ,a man has twenty-five dog kennels all communicating with each other by doorways1 as shown in the illustration4 ,he wishes to arrange his twenty dogs so that they shall form a knight's string from dog ,no4 #a to dog ,no4 #bj1 the bottom row of five kennels to be left empty1 as at present4 ,this is to be done by moving one dog at a time into a vacant kennel4 ,the dogs are well trained to obedience1 and may be trusted to remain in the kennels in which they are placed1 except that if two are placed in the same kennel together they will fight it out to the death4 ,how is the puzzle to be solved in the fewest possible moves without two dogs ever being together8 #cde4--,,the ,,two ,,pawns4 ,illustration ,here is a neat little puzzle in counting4 ,in how many different ways may the two pawns advance to the eighth square8 ,you may move them in any order you like to form a different sequence4 ,for example1 you may move the ,q ,r ,p 7one or two squares7 first1 or the ,k ,r ,p first1 or one pawn as far as you like before touching the other4 ,any sequence is permissible1 only in this puzzle as soon as a pawn reaches the eighth square it is dead1 and remains there unconverted4 ,can you count the number of different sequences8 ,at first it will strike you as being very difficult1 but ,i will show that it is really quite simple when properly attacked4 ,,various ,,chess ,,puzzles4 ,chesse-play is a good and wittie exercise of the minde for some kinde of men4 ,burton's ,anatomy of ,melancholy4 #cdf4--,,setting ,,the ,,board4 ,i have a single chessboard and a single set of chessmen4 ,in how many different ways may the men be correctly set up for the beginning of a game8 ,i find that most people slip at a particular point in making the calculation4 #cdg4--;,,counting ,,the ,,rectangles4 ,can you say correctly just how many squares and other rectangles the chessboard contains8 ,in other words1 in how great a number of different ways is it possible to indicate a square or other rectangle enclosed by lines that separate the squares of the board8 #cdh4--,,the ,,rookery4 ,illustration ,the ,white rooks cannot move outside the little square in which they are enclosed except on the final move1 in giving checkmate4 ,the puzzle is how to checkmate ,black in the fewest possible moves with ,no4 #h rook1 the other rooks being left in numerical order round the sides of their square with the break between #a and #g4 #cdi4--,,stalemate4 ,some years ago the puzzle was proposed to construct an imaginary game of chess1 in which ,white shall be stalemated in the fewest possible moves with all the thirty-two pieces on the board4 ,can you build up such a position in fewer than twenty moves8 #cej4--,,the ,,forsaken ,,king4 ,illustration ,set up the position shown in the diagram4 ,then the condition of the puzzle is--,white to play and checkmate in six moves4 ,notwithstanding the complexities1 ,i will show how the manner of play may be condensed into quite a few lines1 merely stating here that the first two moves of ,white cannot be varied4 #cea4--,,the ,,crusader4 ,the following is a prize puzzle propounded by me some years ago4 ,produce a game of chess which1 after sixteen moves1 shall leave ,white with all his sixteen men on their original squares and ,black in possession of his king alone 7not necessarily on his own square74 ,white is then to force mate in three moves4 #ceb4--;,,immovable ,,pawns4 ,starting from the ordinary arrangement of the pieces as for a game1 what is the smallest possible number of moves necessary in order to arrive at the following position8 ,the moves for both sides must1 of course1 be played strictly in accordance with the rules of the game1 though the result will necessarily be a very weird kind of chess4 ,illustration #cec4--,,thirty-,,six ,,mates4 ,illustration ,place the remaining eight ,white pieces in such a position that ,white shall have the choice of thirty-six different mates on the move4 ,every move that checkmates and leaves a different position is a different mate4 ,the pieces already placed must not be moved4 #ced4--;,,an ,,amazing ,,dilemma4 ,in a game of chess between ,mr4 ,black and ,mr4 ,white1 ,black was in difficulties1 and as usual was obliged to catch a train4 ,so he proposed that ,white should complete the game in his absence on condition that no moves whatever should be made for ,black1 but only with the ,white pieces4 ,mr4 ,white accepted1 but to his dismay found it utterly impossible to win the game under such conditions4 ,try as he would1 he could not checkmate his opponent4 ,on which square did ,mr4 ,black leave his king8 ,the other pieces are in their proper positions in the diagram4 ,white may leave ,black in check as often as he likes1 for it makes no difference1 as he can never arrive at a checkmate position4 ,illustration #cee4--;,,checkmate6 ,illustration ,strolling into one of the rooms of a ,london club1 ,i noticed a position left by two players who had gone4 ,this position is shown in the diagram4 ,it is evident that ,white has checkmated ,black4 ,but how did he do it8 ,that is the puzzle4 #cef4--,,queer ,,chess4 ,can you place two ,white rooks and a ,white knight on the board so that the ,black king 7who must be on one of the four squares in the middle of the board7 shall be in check with no possible move open to him8 ,in other words1 the reader will say1 the king is to be shown checkmated4 ,well1 you can use the term if you wish1 though ,i intentionally do not employ it myself4 ,the mere fact that there is no ,white king on the board would be a sufficient reason for my not doing so4 #ceg4--;,,ancient ,,chinese ,,puzzle4 ,illustration ,my next puzzle is supposed to be ,chinese1 many hundreds of years old1 and never fails to interest4 ,white to play and mate1 moving each of the three pieces once1 and once only4 #ceh4--,,the ,,six ,,pawns4 ,in how many different ways may ,i place six pawns on the chessboard so that there shall be an even number of unoccupied squares in every row and every column8 ,we are not here considering the diagonals at all1 and every different six squares occupied makes a different solution1 so we have not to exclude reversals or reflections4 #cei4--;,,counter ,,solitaire4 ,here is a little game of solitaire that is quite easy1 but not so easy as to be uninteresting4 ,you can either rule out the squares on a sheet of cardboard or paper1 or you can use a portion of your chessboard4 ,i have shown numbered counters in the illustration so as to make the solution easy and intelligible to all1 but chess pawns or draughts will serve just as well in practice4 ,illustration ,the puzzle is to remove all the counters except one1 and this one that is left must be ,no4 #a4 ,you remove a counter by jumping over another counter to the next space beyond1 if that square is vacant1 but you cannot make a leap in a diagonal direction4 ,the following moves will make the play quite clear3 #a-i1 #b-aj1 #a-b1 and so on4 ,here #a jumps over #i1 and you remove #i from the board2 then #b jumps over #aj1 and you remove #aj2 then #a jumps over #b1 and you remove #b4 ,every move is thus a capture1 until the last capture of all is made by ,no4 #a4 #cfj4--;,,chessboard ,,solitaire4 ,illustration ,here is an extension of the last game of solitaire4 ,all you need is a chessboard and the thirty-two pieces1 or the same number of draughts or counters4 ,in the illustration numbered counters are used4 ,the puzzle is to remove all the counters except two1 and these two must have originally been on the same side of the board2 that is1 the two left must either belong to the group #a to #af or to the other group1 #ag to #cb4 ,you remove a counter by jumping over it with another counter to the next square beyond1 if that square is vacant1 but you cannot make a leap in a diagonal direction4 ,the following moves will make the play quite clear3 #c-aa1 #d-ab1 #c-d1 #ac-c4 ,here #c jumps over #aa1 and you remove #aa2 #d jumps over #ab1 and you remove #ab2 and so on4 ,it will be found a fascinating little game of patience1 and the solution requires the exercise of some ingenuity4 #cfa4--,,the ,,monstrosity4 ,one ,christmas ,eve ,i was travelling by rail to a little place in one of the southern counties4 ,the compartment was very full1 and the passengers were wedged in very tightly4 ,my neighbour in one of the corner seats was closely studying a position set up on one of those little folding chessboards that can be carried conveniently in the pocket1 and ,i could scarcely avoid looking at it myself4 ,here is the position3-- ,illustration ,my fellow-passenger suddenly turned his head and caught the look of bewilderment on my face4 ,do you play chess8 he asked4 ,yes1 a little4 ,what is that8 ,a problem8 ,problem8 ,no2 a game4 ,impossible6 ,i exclaimed rather rudely4 ,the position is a perfect monstrosity6 ,he took from his pocket a postcard and handed it to me4 ,it bore an address at one side and on the other the words #dc4 ,k to ,kt #h4 ,it is a correspondence game4 he exclaimed4 ,that is my friend's last move1 and ,i am considering my reply4 ,but you really must excuse me2 the position seems utterly impossible4 ,how on earth1 for example-- ,ah6 he broke in smilingly4 ,i see2 you are a beginner2 you play to win4 ,of course you wouldn't play to lose or draw6 ,he laughed aloud4 ,you have much to learn4 ,my friend and myself do not play for results of that antiquated kind4 ,we seek in chess the wonderful1 the whimsical1 the weird4 ,did you ever see a position like that8 ,i inwardly congratulated myself that ,i never had4 ,that position1 sir1 materializes the sinuous evolvements and syncretic1 synthetic1 and synchronous concatenations of two cerebral individualities4 ,it is the product of an amphoteric and intercalatory interchange of-- ,have you seen the evening paper1 sir8 interrupted the man opposite1 holding out a newspaper4 ,i noticed on the margin beside his thumb some pencilled writing4 ,thanking him1 ,i took the paper and read--,insane1 but quite harmless4 ,he is in my charge4 ,after that ,i let the poor fellow run on in his wild way until both got out at the next station4 ,but that queer position became fixed indelibly in my mind1 with ,black's last move #dc4 ,k to ,kt #h2 and a short time afterwards ,i found it actually possible to arrive at such a position in forty-three moves4 ,can the reader construct such a sequence8 ,how did ,white get his rooks and king's bishop into their present positions1 considering ,black can never have moved his king's bishop8 ,no odds were given1 and every move was perfectly legitimate4 ,,measuring1 ,,weighing1 ,,and ,,packing ,,puzzles4 ,measure still for measure4 ,measure for ,measure1 v4 #a4 ,apparently the first printed puzzle involving the measuring of a given quantity of liquid by pouring from one vessel to others of known capacity was that propounded by ,niccola ,fontana1 better known as ,tartaglia 7the stammerer71 #aejj-aeei4 ,it consists in dividing #bd oz4 of valuable balsam into three equal parts1 the only measures available being vessels holding #e1 #aa1 and #ac ounces respectively4 ,there are many different solutions to this puzzle in six manipulations1 or pourings from one vessel to another4 ,bachet de ,mziriac reprinted this and other of ,tartaglia's puzzles in his ,problmes plaisans et dlectables 7#afab74 ,it is the general opinion that puzzles of this class can only be solved by trial1 but ,i think formul can be constructed for the solution generally of certain related cases4 ,it is a practically unexplored field for investigation4 ,the classic weighing problem is1 of course1 that proposed by ,bachet4 ,it entails the determination of the least number of weights that would serve to weigh any integral number of pounds from #a lb4 to #dj lbs4 inclusive1 when we are allowed to put a weight in either of the two pans4 ,the answer is #a1 #c1 #i1 and #bg lbs4 ,tartaglia had previously propounded the same puzzle with the condition that the weights may only be placed in one pan4 ,the answer in that case is #a1 #b1 #d1 #h1 #af1 #cb lbs4 ,major ,mac,mahon has solved the problem quite generally4 ,a full account will be found in ,ball's ,mathematical ,recreations 7#eth edition74 ,packing puzzles1 in which we are required to pack a maximum number of articles of given dimensions into a box of known dimensions1 are1 ,i believe1 of quite recent introduction4 ,at least ,i cannot recall any example in the books of the old writers4 ,one would rather expect to find in the toy shops the idea presented as a mechanical puzzle1 but ,i do not think ,i have ever seen such a thing4 ,the nearest approach to it would appear to be the puzzles of the jig-saw character1 where there is only one depth of the pieces to be adjusted4 #cfb4--,,the ,,wassail ,,bowl4 ,one ,christmas ,eve three ,weary ,willies came into possession of what was to them a veritable wassail bowl1 in the form of a small barrel1 containing exactly six quarts of fine ale4 ,one of the men possessed a five-pint jug and another a three-pint jug1 and the problem for them was to divide the liquor equally amongst them without waste4 ,of course1 they are not to use any other vessels or measures4 ,if you can show how it was to be done at all1 then try to find the way that requires the fewest possible manipulations1 every separate pouring from one vessel to another1 or down a man's throat1 counting as a manipulation4 #cfc4--,,the ,,doctor',s ,,query4 ,a curious little point occurred to me in my dispensary this morning1 said a doctor4 ,i had a bottle containing ten ounces of spirits of wine1 and another bottle containing ten ounces of water4 ,i poured a quarter of an ounce of spirits into the water and shook them up together4 ,the mixture was then clearly forty to one4 ,then ,i poured back a quarter-ounce of the mixture1 so that the two bottles should again each contain the same quantity of fluid4 ,what proportion of spirits to water did the spirits of wine bottle then contain8 #cfd4--,,the ,,barrel ,,puzzle4 ,the men in the illustration are disputing over the liquid contents of a barrel4 ,what the particular liquid is it is impossible to say1 for we are unable to look into the barrel2 so we will call it water4 ,one man says that the barrel is more than half full1 while the other insists that it is not half full4 ,what is their easiest way of settling the point8 ,it is not necessary to use stick1 string1 or implement of any kind for measuring4 ,i give this merely as one of the simplest possible examples of the value of ordinary sagacity in the solving of puzzles4 ,what are apparently very difficult problems may frequently be solved in a similarly easy manner if we only use a little common sense4 ,illustration #cfe4--,,new ,,measuring ,,puzzle4 ,here is a new poser in measuring liquids that will be found interesting4 ,a man has two ten-quart vessels full of wine1 and a five-quart and a four-quart measure4 ,he wants to put exactly three quarts into each of the two measures4 ,how is he to do it8 ,and how many manipulations 7pourings from one vessel to another7 do you require8 ,of course1 waste of wine1 tilting1 and other tricks are not allowed4 #cff4--,,the ,,honest ,,dairyman4 ,an honest dairyman in preparing his milk for public consumption employed a can marked ,b1 containing milk1 and a can marked ,a1 containing water4 ,from can ,a he poured enough to double the contents of can ,b4 ,then he poured from can ,b into can ,a enough to double its contents4 ,then he finally poured from can ,a into can ,b until their contents were exactly equal4 ,after these operations he would send the can ,a to ,london1 and the puzzle is to discover what are the relative proportions of milk and water that he provides for the ,londoners' breakfast-tables4 ,do they get equal proportions of milk and water--or two parts of milk and one of water--or what8 ,it is an interesting question1 though1 curiously enough1 we are not told how much milk or water he puts into the cans at the start of his operations4 #cfg4--,,wine ,,and ,,water4 ,mr4 ,goodfellow has adopted a capital idea of late4 ,when he gives a little dinner party and the time arrives to smoke1 after the departure of the ladies1 he sometimes finds that the conversation is apt to become too political1 too personal1 too slow1 or too scandalous4 ,then he always manages to introduce to the company some new poser that he has secreted up his sleeve for the occasion4 ,this invariably results in no end of interesting discussion and debate1 and puts everybody in a good humour4 ,here is a little puzzle that he propounded the other night1 and it is extraordinary how the company differed in their answers4 ,he filled a wine-glass half full of wine1 and another glass twice the size one-third full of wine4 ,then he filled up each glass with water and emptied the contents of both into a tumbler4 ,now1 he said1 what part of the mixture is wine and what part water8 ,can you give the correct answer8 #cfh4--,,the ,,keg ,,of ,,wine4 ,here is a curious little problem4 ,a man had a ten-gallon keg full of wine and a jug4 ,one day he drew off a jugful of wine and filled up the keg with water4 ,later on1 when the wine and water had got thoroughly mixed1 he drew off another jugful and again filled up the keg with water4 ,it was then found that the keg contained equal proportions of wine and water4 ,can you find from these facts the capacity of the jug8 #cfi4--,,mixing ,,the ,,tea4 ,mrs4 ,spooner called this morning1 said the honest grocer to his assistant4 ,she wants twenty pounds of tea at #bs4 #d;d4 per lb4 ,of course we have a good #bs4 #f;d4 tea1 a slightly inferior at #bs4 #c;d41 and a cheap ,indian at #as4 #i;d41 but she is very particular always about her prices4 ,what do you propose to do8 asked the innocent assistant4 ,do8 exclaimed the grocer4 ,why1 just mix up the three teas in different proportions so that the twenty pounds will work out fairly at the lady's price4 ,only don't put in more of the best tea than you can help1 as we make less profit on that1 and of course you will use only our complete pound packets4 ,don't do any weighing4 ,how was the poor fellow to mix the three teas8 ,could you have shown him how to do it8 #cgj4--;,a ,,packing ,,puzzle4 ,as we all know by experience1 considerable ingenuity is often required in packing articles into a box if space is not to be unduly wasted4 ,a man once told me that he had a large number of iron balls1 all exactly two inches in diameter1 and he wished to pack as many of these as possible into a rectangular box #bdi_/aj inches long1 #bbd_/e inches wide1 and #ad inches deep4 ,now1 what is the greatest number of the balls that he could pack into that box8 #cga4--;,,gold ,,packing ,,in ,,russia4 ,the editor of the ,times newspaper was invited by a high ,russian official to inspect the gold stored in reserve at ,st4 ,petersburg1 in order that he might satisfy himself that it was not another ,humbert safe4 ,he replied that it would be of no use whatever1 for although the gold might appear to be there1 he would be quite unable from a mere inspection to declare that what he saw was really gold4 ,a correspondent of the ,daily ,mail thereupon took up the challenge1 but1 although he was greatly impressed by what he saw1 he was compelled to confess his incompetence 7without emptying and counting the contents of every box and sack1 and assaying every piece of gold7 to give any assurance on the subject4 ,in presenting the following little puzzle1 ,i wish it to be also understood that ,i do not guarantee the real existence of the gold1 and the point is not at all material to our purpose4 ,moreover1 if the reader says that gold is not usually put up in slabs of the dimensions that ,i give1 ,i can only claim problematic licence4 ,russian officials were engaged in packing #hjj gold slabs1 each measuring #ab inches long1 #aa inches wide1 and #a inch deep4 ,what are the interior dimensions of a box of equal length and width1 and necessary depth1 that will exactly contain them without any space being left over8 ,not more than twelve slabs may be laid on edge1 according to the rules of the government4 ,it is an interesting little problem in packing1 and not at all difficult4 #cgb4--,,the ,,barrels ,,of ,,honey4 ,illustration ,once upon a time there was an aged merchant of ,bagdad who was much respected by all who knew him4 ,he had three sons1 and it was a rule of his life to treat them all exactly alike4 ,whenever one received a present1 the other two were each given one of equal value4 ,one day this worthy man fell sick and died1 bequeathing all his possessions to his three sons in equal shares4 ,the only difficulty that arose was over the stock of honey4 ,there were exactly twenty-one barrels4 ,the old man had left instructions that not only should every son receive an equal quantity of honey1 but should receive exactly the same number of barrels1 and that no honey should be transferred from barrel to barrel on account of the waste involved4 ,now1 as seven of these barrels were full of honey1 seven were half-full1 and seven were empty1 this was found to be quite a puzzle1 especially as each brother objected to taking more than four barrels of1 the same description--full1 half-full1 or empty4 ,can you show how they succeeded in making a correct division of the property8 ,,crossing ,,river ,,problems ,my boat is on the shore4 ,,byron4 ,this is another medival class of puzzles4 ,probably the earliest example was by ,abbot ,alcuin1 who was born in ,yorkshire in #gce and died at ,tours in #hjd4 ,and everybody knows the story of the man with the wolf1 goat1 and basket of cabbages whose boat would only take one of the three at a time with the man himself4 ,his difficulties arose from his being unable to leave the wolf alone with the goat1 or the goat alone with the cabbages4 ,these puzzles were considered by ,tartaglia and ,bachet1 and have been later investigated by ,lucas1 ,de ,fonteney1 ,delannoy1 ,tarry1 and others4 ,in the puzzles ,i give there will be found one or two new conditions which add to the complexity somewhat4 ,i also include a pulley problem that practically involves the same principles4 ,illustration #cgc4--;,,crossing ,,the ,,stream4 ,during a country ramble ,mr4 and ,mrs4 ,softleigh found themselves in a pretty little dilemma4 ,they had to cross a stream in a small boat which was capable of carrying only #aej lbs4 weight4 ,but ,mr4 ,softleigh and his wife each weighed exactly #aej lbs41 and each of their sons weighed #ge lbs4 ,and then there was the dog1 who could not be induced on any terms to swim4 ,on the principle of ladies first1 they at once sent ,mrs4 ,softleigh over2 but this was a stupid oversight1 because she had to come back again with the boat1 so nothing was gained by that operation4 ,how did they all succeed in getting across8 ,the reader will find it much easier than the ,softleigh family did1 for their greatest enemy could not have truthfully called them a brilliant quartette--while the dog was a perfect fool4 #cgd--;,,crossing ,,the ,,river ,,axe4 ,many years ago1 in the days of the smuggler known as ,rob ,roy of the ,west1 a piratical band buried on the coast of ,south ,devon a quantity of treasure which was1 of course1 abandoned by them in the usual inexplicable way4 ,some time afterwards its whereabouts was discovered by three countrymen1 who visited the spot one night and divided the spoil between them1 ,giles taking treasure to the value of #hjj1 ,jasper #ejj worth1 and ,timothy #cjj worth4 ,in returning they had to cross the river ,axe at a point where they had left a small boat in readiness4 ,here1 however1 was a difficulty they had not anticipated4 ,the boat would only carry two men1 or one man and a sack1 and they had so little confidence in one another that no person could be left alone on the land or in the boat with more than his share of the spoil1 though two persons 7being a check on each other7 might be left with more than their shares4 ,the puzzle is to show how they got over the river in the fewest possible crossings1 taking their treasure with them4 ,no tricks1 such as ropes1 flying bridges1 currents1 swimming1 or similar dodges1 may be employed4 #cge4--;,,five ,,jealous ,,husbands4 ,during certain local floods five married couples found themselves surrounded by water1 and had to escape from their unpleasant position in a boat that would only hold three persons at a time4 ,every husband was so jealous that he would not allow his wife to be in the boat or on either bank with another man 7or with other men7 unless he was himself present4 ,show the quickest way of getting these five men and their wives across into safety4 ,call the men ,a1 ,b1 ,c1 ,d1 ,e1 and their respective wives a1 b1 c1 d1 e4 ,to go over and return counts as two crossings4 ,no tricks such as ropes1 swimming1 currents1 etc41 are permitted4 #cgf4--,,the ,,four ,,elopements4 ,colonel ,b---- was a widower of a very taciturn disposition4 ,his treatment of his four daughters was unusually severe1 almost cruel1 and they not unnaturally felt disposed to resent it4 ,being charming girls with every virtue and many accomplishments1 it is not surprising that each had a fond admirer4 ,but the father forbade the young men to call at his house1 intercepted all letters1 and placed his daughters under stricter supervision than ever4 ,but love1 which scorns locks and keys and garden walls1 was equal to the occasion1 and the four youths conspired together and planned a general elopement4 ,at the foot of the tennis lawn at the bottom of the garden ran the silver ,thames1 and one night1 after the four girls had been safely conducted from a dormitory window to terra firma1 they all crept softly down to the bank of the river1 where a small boat belonging to the ,colonel was moored4 ,with this they proposed to cross to the opposite side and make their way to a lane where conveyances were waiting to carry them in their flight4 ,alas6 here at the water's brink their difficulties already began4 ,the young men were so extremely jealous that not one of them would allow his prospective bride to remain at any time in the company of another man1 or men1 unless he himself were present also4 ,now1 the boat would only hold two persons1 though it could1 of course1 be rowed by one1 and it seemed impossible that the four couples would ever get across4 ,but midway in the stream was a small island1 and this seemed to present a way out of the difficulty1 because a person or persons could be left there while the boat was rowed back or to the opposite shore4 ,if they had been prepared for their difficulty they could have easily worked out a solution to the little poser at any other time4 ,but they were now so hurried and excited in their flight that the confusion they soon got into was exceedingly amusing--or would have been to any one except themselves4 ,as a consequence they took twice as long and crossed the river twice as often as was really necessary4 ,meanwhile1 the ,colonel1 who was a very light sleeper1 thought he heard a splash of oars4 ,he quickly raised the alarm among his household1 and the young ladies were found to be missing4 ,somebody was sent to the police-station1 and a number of officers soon aided in the pursuit of the fugitives1 who1 in consequence of that delay in crossing the river1 were quickly overtaken4 ,the four girls returned sadly to their homes1 and afterwards broke off their engagements in disgust4 ,for a considerable time it was a mystery how the party of eight managed to cross the river in that little boat without any girl being ever left with a man1 unless her betrothed was also present4 ,the favourite method is to take eight counters or pieces of cardboard and mark them ,a1 ,b1 ,c1 ,d1 a1 b1 c1 d1 to represent the four men and their prospective brides1 and carry them from one side of a table to the other in a matchbox 7to represent the boat71 a penny being placed in the middle of the table as the island4 ,readers are now asked to find the quickest method of getting the party across the river4 ,how many passages are necessary from land to land8 ,by land is understood either shore or island4 ,though the boat would not necessarily call at the island every time of crossing1 the possibility of its doing so must be provided for4 ,for example1 it would not do for a man to be alone in the boat 7though it were understood that he intended merely to cross from one bank to the opposite one7 if there happened to be a girl alone on the island other than the one to whom he was engaged4 #cgg4--,,stealing ,,the ,,castle ,,treasure4 ,the ingenious manner in which a box of treasure1 consisting principally of jewels and precious stones1 was stolen from ,gloomhurst ,castle has been handed down as a tradition in the ,de ,gourney family4 ,the thieves consisted of a man1 a youth1 and a small boy1 whose only mode of escape with the box of treasure was by means of a high window4 ,outside the window was fixed a pulley1 over which ran a rope with a basket at each end4 ,when one basket was on the ground the other was at the window4 ,the rope was so disposed that the persons in the basket could neither help themselves by means of it nor receive help from others4 ,in short1 the only way the baskets could be used was by placing a heavier weight in one than in the other4 ,now1 the man weighed #aie lbs41 the youth #aje lbs41 the boy #ij lbs41 and the box of treasure #ge lbs4 ,the weight in the descending basket could not exceed that in the other by more than #ae lbs4 without causing a descent so rapid as to be most dangerous to a human being1 though it would not injure the stolen property4 ,only two persons1 or one person and the treasure1 could be placed in the same basket at one time4 ,how did they all manage to escape and take the box of treasure with them8 ,the puzzle is to find the shortest way of performing the feat1 which in itself is not difficult4 ,remember1 a person cannot help himself by hanging on to the rope1 the only way being to go down with a bump1 with the weight in the other basket as a counterpoise4 ,,problems ,,concerning ,,games4 ,the little pleasure of the game4 ,,matthew ,,prior4 ,every game lends itself to the propounding of a variety of puzzles4 ,they can be made1 as we have seen1 out of the chessboard and the peculiar moves of the chess pieces4 ,i will now give just a few examples of puzzles with playing cards and dominoes1 and also go out of doors and consider one or two little posers in the cricket field1 at the football match1 and the horse race and motor-car race4 #cgh4--;,,dominoes ,,in ,,progression4 ,illustration ,it will be seen that ,i have played six dominoes1 in the illustration1 in accordance with the ordinary rules of the game1 #d against #d1 #a against #a1 and so on1 and yet the sum of the spots on the successive dominoes1 #d1 #e1 #f1 #g1 #h1 #i1 are in arithmetical progression2 that is1 the numbers taken in order have a common difference of #a4 ,in how many different ways may we play six dominoes1 from an ordinary box of twenty-eight1 so that the numbers on them may lie in arithmetical progression8 ,we must always play from left to right1 and numbers in decreasing arithmetical progression 7such as #i1 #h1 #g1 #f1 #e1 #d7 are not admissible4 #cgi4--,,the ,,five ,,dominoes4 ,illustration ,here is a new little puzzle that is not difficult1 but will probably be found entertaining by my readers4 ,it will be seen that the five dominoes are so arranged in proper sequence 7that is1 with #a against #a1 #b against #b1 and so on71 that the total number of pips on the two end dominoes is five1 and the sum of the pips on the three dominoes in the middle is also five4 ,there are just three other arrangements giving five for the additions4 ,they are3 -- 7#a--j7 7#j--j7 7#j--b7 7#b--a7 7#a--c7 7#d--j7 7#j--j7 7#j--b7 7#b--a7 7#a--j7 7#b--j7 7#j--j7 7#j--a7 7#a--c7 7#c--j7 ,now1 how many similar arrangements are there of five dominoes that shall give six instead of five in the two additions8 #chj4--,,the ,,domino ,,frame ,,puzzle4 ,illustration ,it will be seen in the illustration that the full set of twenty-eight dominoes is arranged in the form of a square frame1 with #f against #f1 #b against #b1 blank against blank1 and so on1 as in the game4 ,it will be found that the pips in the top row and left-hand column both add up #dd4 ,the pips in the other two sides sum to #ei and #cb respectively4 ,the puzzle is to rearrange the dominoes in the same form so that all of the four sides shall sum to #dd4 ,remember that the dominoes must be correctly placed one against another as in the game4 #cha4--,,the ,,card ,,frame ,,puzzle4 ,in the illustration we have a frame constructed from the ten playing cards1 ace to ten of diamonds4 ,the children who made it wanted the pips on all four sides to add up alike1 but they failed in their attempt and gave it up as impossible4 ,it will be seen that the pips in the top row1 the bottom row1 and the left-hand side all add up #ad1 but the right-hand side sums to #bc4 ,now1 what they were trying to do is quite possible4 ,can you rearrange the ten cards in the same formation so that all four sides shall add up alike8 ,of course they need not add up #ad1 but any number you choose to select4 ,illustration #chb4--,,the ,,cross ,,of ,,cards4 ,illustration ,in this case we use only nine cards--the ace to nine of diamonds4 ,the puzzle is to arrange them in the form of a cross1 exactly in the way shown in the illustration1 so that the pips in the vertical bar and in the horizontal bar add up alike4 ,in the example given it will be found that both directions add up #bc4 ,what ,i want to know is1 how many different ways are there of rearranging the cards in order to bring about this result8 ,it will be seen that1 without affecting the solution1 we may exchange the #e with the #f1 the #e with the #g1 the #h with the #c1 and so on4 ,also we may make the horizontal and the vertical bars change places4 ,but such obvious manipulations as these are not to be regarded as different solutions4 ,they are all mere variations of one fundamental solution4 ,now1 how many of these fundamentally different solutions are there8 ,the pips need not1 of course1 always add up #bc4 #chc4--,,the ,t ,,card ,,puzzle4 ,illustration ,an entertaining little puzzle with cards is to take the nine cards of a suit1 from ace to nine inclusive1 and arrange them in the form of the letter ,t1 as shown in the illustration1 so that the pips in the horizontal line shall count the same as those in the column4 ,in the example given they add up twenty-three both ways4 ,now1 it is quite easy to get a single correct arrangement4 ,the puzzle is to discover in just how many different ways it may be done4 ,though the number is high1 the solution is not really difficult if we attack the puzzle in the right manner4 ,the reverse way obtained by reflecting the illustration in a mirror we will not count as different1 but all other changes in the relative positions of the cards will here count4 ,how many different ways are there8 #chd4--;,,card ,,triangles4 ,here you pick out the nine cards1 ace to nine of diamonds1 and arrange them in the form of a triangle1 exactly as shown in the illustration1 so that the pips add up the same on the three sides4 ,in the example given it will be seen that they sum to #bj on each side1 but the particular number is of no importance so long as it is the same on all three sides4 ,the puzzle is to find out in just how many different ways this can be done4 ,if you simply turn the cards round so that one of the other two sides is nearest to you this will not count as different1 for the order will be the same4 ,also1 if you make the #d1 #i1 #e change places with the #g1 #c1 #h1 and at the same time exchange the #a and the #f1 it will not be different4 ,but if you only change the #a and the #f it will be different1 because the order round the triangle is not the same4 ,this explanation will prevent any doubt arising as to the conditions4 ,illustration #che4--,,strand ,,patience4 ,the idea for this came to me when considering the game of ,patience that ,i gave in the ,strand ,magazine for ,december1 #aiaj1 which has been reprinted in ,ernest ,bergholt's ,second ,book of ,patience ,games1 under the new name of ,king ,albert4 ,make two piles of cards as follows3 #i ,d1 #h ,s1 #g ,d1 #f ,s1 #e ,d1 #d ,s1 #c ,d1 #b ,s1 #a ,d1 and #i ,h1 #h ,c1 #g ,h1 #f ,c1 #e ,h1 #d ,c1 #c ,h1 #b ,c1 #a ,h1 with the #i of diamonds at the bottom of one pile and the #i of hearts at the bottom of the other4 ,the point is to exchange the spades with the clubs1 so that the diamonds and clubs are still in numerical order in one pile and the hearts and spades in the other4 ,there are four vacant spaces in addition to the two spaces occupied by the piles1 and any card may be laid on a space1 but a card can only be laid on another of the next higher value--an ace on a two1 a two on a three1 and so on4 ,patience is required to discover the shortest way of doing this4 ,when there are four vacant spaces you can pile four cards in seven moves1 with only three spaces you can pile them in nine moves1 and with two spaces you cannot pile more than two cards4 ,when you have a grasp of these and similar facts you will be able to remove a number of cards bodily and write down #g1 #i1 or whatever the number of moves may be4 ,the gradual shortening of play is fascinating1 and first attempts are surprisingly lengthy4 #chf4--;,a ,,trick ,,with ,,dice4 ,illustration ,here is a neat little trick with three dice4 ,i ask you to throw the dice without my seeing them4 ,then ,i tell you to multiply the points of the first die by #b and add #e2 then multiply the result by #e and add the points of the second die2 then multiply the result by #aj and add the points of the third die4 ,you then give me the total1 and ,i can at once tell you the points thrown with the three dice4 ,how do ,i do it8 ,as an example1 if you threw #a1 #c1 and #f1 as in the illustration1 the result you would give me would be #chf1 from which ,i could at once say what you had thrown4 #chg4--,,the ,,village ,,cricket ,,match4 ,in a cricket match1 ,dingley ,dell v4 ,all ,muggleton1 the latter had the first innings4 ,mr4 ,dumkins and ,mr4 ,podder were at the wickets1 when the wary ,dumkins made a splendid late cut1 and ,mr4 ,podder called on him to run4 ,four runs were apparently completed1 but the vigilant umpires at each end called1 three short1 making six short runs in all4 ,what number did ,mr4 ,dumkins score8 ,when ,dingley ,dell took their turn at the wickets their champions were ,mr4 ,luffey and ,mr4 ,struggles4 ,the latter made a magnificent off-drive1 and invited his colleague to come along1 with the result that the observant spectators applauded them for what was supposed to have been three sharp runs4 ,but the umpires declared that there had been two short runs at each end--four in all4 ,to what extent1 if any1 did this manoeuvre increase ,mr4 ,struggles's total8 #chh4--,,slow ,,cricket4 ,in the recent county match between ,wessex and ,nincomshire the former team were at the wickets all day1 the last man being put out a few minutes before the time for drawing stumps4 ,the play was so slow that most of the spectators were fast asleep1 and1 on being awakened by one of the officials clearing the ground1 we learnt that two men had been put out leg-before-wicket for a combined score of #ai runs2 four men were caught for a combined score or #ag runs2 one man was run out for a duck's egg2 and the others were all bowled for #c runs each4 ,there were no extras4 ,we were not told which of the men was the captain1 but he made exactly #ae more than the average of his team4 ,what was the captain's score8 #chi4--,,the ,,football ,,players4 ,it is a glorious game6 an enthusiast was heard to exclaim4 ,at the close of last season1 of the footballers of my acquaintance four had broken their left arm1 five had broken their right arm1 two had the right arm sound1 and three had sound left arms4 ,can you discover from that statement what is the smallest number of players that the speaker could be acquainted with8 ,it does not at all follow that there were as many as fourteen men1 because1 for example1 two of the men who had broken the left arm might also be the two who had sound right arms4 #cij4--,,the ,,horse-,,race ,,puzzle4 ,there are no morals in puzzles4 ,when we are solving the old puzzle of the captain who1 having to throw half his crew overboard in a storm1 arranged to draw lots1 but so placed the men that only the ,turks were sacrificed1 and all the ,christians left on board1 we do not stop to discuss the questionable morality of the proceeding4 ,and when we are dealing with a measuring problem1 in which certain thirsty pilgrims are to make an equitable division of a barrel of beer1 we do not object that1 as total abstainers1 it is against our conscience to have anything to do with intoxicating liquor4 ,therefore ,i make no apology for introducing a puzzle that deals with betting4 ,three horses--,acorn1 ,bluebottle1 and ,capsule--start in a race4 ,the odds are #d to #a1 ,acorn2 #c to #a1 ,bluebottle2 #b to #a1 ,capsule4 ,now1 how much must ,i invest on each horse in order to win #ac1 no matter which horse comes in first8 ,supposing1 as an example1 that ,i betted #e on each horse4 ,then1 if ,acorn won1 ,i should receive #bj 7four times #e71 and have to pay #e each for the other two horses2 thereby winning #aj4 ,but it will be found that if ,bluebottle was first ,i should only win #e1 and if ,capsule won ,i should gain nothing and lose nothing4 ,this will make the question perfectly clear to the novice1 who1 like myself1 is not interested in the calling of the fraternity who profess to be engaged in the noble task of improving the breed of horses4 #cia4--,,the ,,motor-,,car ,,race4 ,sometimes a quite simple statement of fact1 if worded in an unfamiliar manner1 will cause considerable perplexity4 ,here is an example1 and it will doubtless puzzle some of my more youthful readers just a little4 ,i happened to be at a motor-car race at ,brooklands1 when one spectator said to another1 while a number of cars were whirling round and round the circular track3-- ,there's ,gogglesmith--that man in the white car6 ,yes1 ,i see1 was the reply2 but how many cars are running in this race8 ,then came this curious rejoinder3-- ,one-third of the cars in front of ,gogglesmith added to three-quarters of those behind him will give you the answer4 ,now1 can you tell how many cars were running in the race8 ,,puzzle ,,games4 ,he that is beaten may be said ,to lie in honour's truckle bed4 ,,hudibras4 ,it may be said generally that a game is a contest of skill for two or more persons1 into which we enter either for amusement or to win a prize4 ,a puzzle is something to be done or solved by the individual4 ,for example1 if it were possible for us so to master the complexities of the game of chess that we could be assured of always winning with the first or second move1 as the case might be1 or of always drawing1 then it would cease to be a game and would become a puzzle4 ,of course among the young and uninformed1 when the correct winning play is not understood1 a puzzle may well make a very good game4 ,thus there is no doubt children will continue to play ,noughts and ,crosses1 though ,i have shown 7,no4 #aji1 ,canterbury ,puzzles7 that between two players who both thoroughly understand the play1 every game should be drawn4 ,neither player could ever win except through the blundering of his opponent4 ,but ,i am writing from the point of view of the student of these things4 ,the examples that ,i give in this class are apparently games1 but1 since ,i show in every case how one player may win if he only play correctly1 they are in reality puzzles4 ,their interest1 therefore1 lies in attempting to discover the leading method of play4 #cib4--,,the ,,pebble ,,game4 ,here is an interesting little puzzle game that ,i used to play with an acquaintance on the beach at ,slocomb-on-,sea4 ,two players place an odd number of pebbles1 we will say fifteen1 between them4 ,then each takes in turn one1 two1 or three pebbles 7as he chooses71 and the winner is the one who gets the odd number4 ,thus1 if you get seven and your opponent eight1 you win4 ,if you get six and he gets nine1 he wins4 ,ought the first or second player to win1 and how8 ,when you have settled the question with fifteen pebbles try again with1 say1 thirteen4 #cic4--,,the ,,two ,,rooks4 ,this is a puzzle game for two players4 ,each player has a single rook4 ,the first player places his rook on any square of the board that he may choose to select1 and then the second player does the same4 ,they now play in turn1 the point of each play being to capture the opponent's rook4 ,but in this game you cannot play through a line of attack without being captured4 ,that is to say1 if in the diagram it is ,black's turn to play1 he cannot move his rook to his king's knight's square1 or to his king's rook's square1 because he would enter the line of fire when passing his king's bishop's square4 ,for the same reason he cannot move to his queen's rook's seventh or eighth squares4 ,now1 the game can never end in a draw4 ,sooner or later one of the rooks must fall1 unless1 of course1 both players commit the absurdity of not trying to win4 ,the trick of winning is ridiculously simple when you know it4 ,can you solve the puzzle8 ,illustration #cid4--,,puss ,,in ,,the ,,corner4 ,illustration ,this variation of the last puzzle is also played by two persons4 ,one puts a counter on ,no4 #f1 and the other puts one on ,no4 #ee1 and they play alternately by removing the counter to any other number in a line4 ,if your opponent moves at any time on to one of the lines you occupy1 or even crosses one of your lines1 you immediately capture him and win4 ,we will take an illustrative game4 ,a moves from #ee to #eb2 ,b moves from #f to #ac2 ,a advances to #bc2 ,b goes to #ae2 ,a retreats to #bf2 ,b retreats to #ac2 ,a advances to #ba2 ,b retreats to #b2 ,a advances to #g2 ,b goes to #c2 ,a moves to #f2 ,b must now go to #d2 ,a establishes himself at #aa1 and ,b must be captured next move because he is compelled to cross a line on which ,a stands4 ,play this over and you will understand the game directly4 ,now1 the puzzle part of the game is this3 ,which player should win1 and how many moves are necessary8 #cie4--;,a ,,war ,,puzzle ,,game4 ,illustration ,here is another puzzle game4 ,one player1 representing the ,british general1 places a counter at ,b1 and the other player1 representing the enemy1 places his counter at ,e4 ,the ,britisher makes the first advance along one of the roads to the next town1 then the enemy moves to one of his nearest towns1 and so on in turns1 until the ,british general gets into the same town as the enemy and captures him4 ,although each must always move along a road to the next town only1 and the second player may do his utmost to avoid capture1 the ,british general 7as we should suppose1 from the analogy of real life7 must infallibly win4 ,but how8 ,that is the question4 #cif4--;,a ,,match ,,mystery4 ,here is a little game that is childishly simple in its conditions4 ,but it is well worth investigation4 ,mr4 ,stubbs pulled a small table between himself and his friend1 ,mr4 ,wilson1 and took a box of matches1 from which he counted out thirty4 ,here are thirty matches1 he said4 ,i divide them into three unequal heaps4 ,let me see4 ,we have #ad1 #aa1 and #e1 as it happens4 ,now1 the two players draw alternately any number from any one heap1 and he who draws the last match loses the game4 ,that's all6 ,i will play with you1 ,wilson4 ,i have formed the heaps1 so you have the first draw4 ,as ,i can draw any number1 ,mr4 ,wilson said1 suppose ,i exhibit my usual moderation and take all the #ad heap4 ,that is the worst you could do1 for it loses right away4 ,i take #f from the #aa1 leaving two equal heaps of #e1 and to leave two equal heaps is a certain win 7with the single exception of #a1 #a71 because whatever you do in one heap ,i can repeat in the other4 ,if you leave #d in one heap1 ,i leave #d in the other4 ,if you then leave #b in one heap1 ,i leave #b in the other4 ,if you leave only #a in one heap1 then ,i take all the other heap4 ,if you take all one heap1 ,i take all but one in the other4 ,no1 you must never leave two heaps1 unless they are equal heaps and more than #a1 #a4 ,let's begin again4 ,very well1 then1 said ,mr4 ,wilson4 ,i will take #f from the #ad1 and leave you #h1 #aa1 #e4 ,mr4 ,stubbs then left #h1 #aa1 #c2 ,mr4 ,wilson1 #h1 #e1 #c2 ,mr4 ,stubbs1 #f1 #e1 #c2 ,mr4 ,wilson1#d1 #e1 #c2 ,mr4 ,stubbs1 #d1 #e1 #a2 ,mr4 ,wilson1 #d1 #c1 #a2 ,mr4 ,stubbs1 #b1 #c1 #a2 ,mr4 ,wilson1 #b1 #a1 #a2 which ,mr4 ,stubbs reduced to #a1 #a1 #a4 ,it is now quite clear that ,i must win1 said ,mr4 ,stubbs1 because you must take #a1 and then ,i take #a1 leaving you the last match4 ,you never had a chance4 ,there are just thirteen different ways in which the matches may be grouped at the start for a certain win4 ,in fact1 the groups selected1 #ad1 #aa1 #e1 are a certain win1 because for whatever your opponent may play there is another winning group you can secure1 and so on and on down to the last match4 #cig4--,,the ,,montenegrin ,,dice ,,game4 ,it is said that the inhabitants of ,montenegro have a little dice game that is both ingenious and well worth investigation4 ,the two players first select two different pairs of odd numbers 7always higher than #c7 and then alternately toss three dice4 ,whichever first throws the dice so that they add up to one of his selected numbers wins4 ,if they are both successful in two successive throws it is a draw and they try again4 ,for example1 one player may select #g and #ae and the other #e and #ac4 ,then if the first player throws so that the three dice add up #g or #ae he wins1 unless the second man gets either #e or #ac on his throw4 ,the puzzle is to discover which two pairs of numbers should be selected in order to give both players an exactly even chance4 #cih4--,,the ,,cigar ,,puzzle4 ,i once propounded the following puzzle in a ,london club1 and for a considerable period it absorbed the attention of the members4 ,they could make nothing of it1 and considered it quite impossible of solution4 ,and yet1 as ,i shall show1 the answer is remarkably simple4 ,two men are seated at a square-topped table4 ,one places an ordinary cigar 7flat at one end1 pointed at the other7 on the table1 then the other does the same1 and so on alternately1 a condition being that no cigar shall touch another4 ,which player should succeed in placing the last cigar1 assuming that they each will play in the best possible manner8 ,the size of the table top and the size of the cigar are not given1 but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar1 we will say that the table must not be less than #b feet square and the cigar not more than #d inches long4 ,with those restrictions you may take any dimensions you like4 ,of course we assume that all the cigars are exactly alike in every respect4 ,should the first player1 or the second player1 win8 ,,magic ,,square ,,problems4 ,by magic numbers4 ,,congreve1 ,the ,mourning ,bride4 ,this is a very ancient branch of mathematical puzzledom1 and it has an immense1 though scattered1 literature of its own4 ,in their simple form of consecutive whole numbers arranged in a square so that every column1 every row1 and each of the two long diagonals shall add up alike1 these magic squares offer three main lines of investigation3 ,construction1 ,enumeration1 and ,classification4 ,of recent years many ingenious methods have been devised for the construction of magics1 and the law of their formation is so well understood that all the ancient mystery has evaporated and there is no longer any difficulty in making squares of any dimensions4 ,almost the last word has been said on this subject4 ,the question of the enumeration of all the possible squares of a given order stands just where it did over two hundred years ago4 ,everybody knows that there is only one solution for the third order1 three cells by three2 and ,frnicle published in #afic diagrams of all the arrangements of the fourth order--#hhj in number--and his results have been verified over and over again4 ,i may here refer to the general solution for this order1 for numbers not necessarily consecutive1 by ,e4 ,bergholt in ,nature1 ,may #bf1 #aiaj1 as it is of the greatest importance to students of this subject4 ,the enumeration of the examples of any higher order is a completely unsolved problem4 ,as to classification1 it is largely a matter of individual taste--perhaps an sthetic question1 for there is beauty in the law and order of numbers4 ,a man once said that he divided the human race into two great classes3 those who take snuff and those who do not4 ,i am not sure that some of our classifications of magic squares are not almost as valueless4 ,however1 lovers of these things seem somewhat agreed that ,nasik magic squares 7so named by ,mr4 ,frost1 a student of them1 after the town in ,india where he lived1 and also called ,diabolique and ,pandiagonal7 and ,associated magic squares are of special interest1 so ,i will just explain what these are for the benefit of the novice4 ,illustration3 ,,simple ,illustration3 ,,semi-,,nasik ,illustration3 ,,associated ,illustration3 ,,nasik ,i published in ,the ,queen for ,january #ae1 #aiaj1 an article that would enable the reader to write out1 if he so desired1 all the #hhj magics of the fourth order1 and the following is the complete classification that ,i gave4 ,the first example is that of a ,simple square that fulfils the simple conditions and no more4 ,the second example is a ,semi-,nasik1 which has the additional property that the opposite short diagonals of two cells each together sum to #cd4 ,thus1 #ad #d #aa #e #cd and #ab #f #ac #c #cd4 ,the third example is not only ,semi-,nasik but also ,associated1 because in it every number1 if added to the number that is equidistant1 in a straight line1 from the centre gives #ag4 ,thus1 #a #af1 #b #ae1 #c #ad1 etc4 ,the fourth example1 considered the most perfect of all1 is a ,nasik4 ,here all the broken diagonals sum to #cd4 ,thus1 for example1 #ae #ad #b #c1 and #aj #d #g #ac1 and #ae #e #b #ab4 ,as a consequence1 its properties are such that if you repeat the square in all directions you may mark off a square1 #d #d1 wherever you please1 and it will be magic4 ,the following table not only gives a complete enumeration under the four forms described1 but also a classification under the twelve graphic types indicated in the diagrams4 ,the dots at the end of each line represent the relative positions of those complementary pairs1 #a #af1 #b #ae1 etc41 which sum to #ag4 ,for example1 it will be seen that the first and second magic squares given are of ,type ,,vi41 that the third square is of ,type ,,iii41 and that the fourth is of ,type ,i4 ,edouard ,lucas indicated these types1 but he dropped exactly half of them and did not attempt the classification4 ,,nasik 7,type ,i47 4 4 4 4 4 #dh ,,semi-,,nasik 7,type ,,ii41 ,transpositions of ,nasik7 4 #dh 7,type ,,iii41 ,associated7 #dh 7,type ,,iv47 4 4 4 #if 7,type ,v47 4 4 4 #if #aib 7,type ,,vi47 4 4 4 #if #chd ,,simple4 7,type ,,vi47 4 4 4 #bjh 7,type ,,vii47 4 4 4 #ef 7,type ,,viii474 4 4 #ef 7,type ,,ix47 4 4 4 #ef 7,type ,x47 4 4 4 #ef #bbd 7,type ,,xi47 4 4 4 #h 7,type ,,xii47 4 4 4 #h #af #ddh #hhj ,it is hardly necessary to say that every one of these squares will produce seven others by mere reversals and reflections1 which we do not count as different4 ,so that there are #g1jdj squares of this order1 #hhj of which are fundamentally different4 ,an infinite variety of puzzles may be made introducing new conditions into the magic square4 ,in ,the ,canterbury ,puzzles ,i have given examples of such squares with coins1 with postage stamps1 with cutting-out conditions1 and other tricks4 ,i will now give a few variants involving further novel conditions4 #cii4--,,the ,,troublesome ,,eight4 ,nearly everybody knows that a magic square is an arrangement of numbers in the form of a square so that every row1 every column1 and each of the two long diagonals adds up alike4 ,for example1 you would find little difficulty in merely placing a different number in each of the nine cells in the illustration so that the rows1 columns1 and diagonals shall all add up #ae4 ,and at your first attempt you will probably find that you have an #h in one of the corners4 ,the puzzle is to construct the magic square1 under the same conditions1 with the #h in the position shown4 ,illustration #djj4--,,the ,,magic ,,strips4 ,illustration ,i happened to have lying on my table a number of strips of cardboard1 with numbers printed on them from #a upwards in numerical order4 ,the idea suddenly came to me1 as ideas have a way of unexpectedly coming1 to make a little puzzle of this4 ,i wonder whether many readers will arrive at the same solution that ,i did4 ,take seven strips of cardboard and lay them together as above4 ,then write on each of them the numbers #a1 #b1 #c1 #d1 #e1 #f1 #g1 as shown1 so that the numbers shall form seven rows and seven columns4 ,now1 the puzzle is to cut these strips into the fewest possible pieces so that they may be placed together and form a magic square1 the seven rows1 seven columns1 and two diagonals adding up the same number4 ,no figures may be turned upside down or placed on their sides--that is1 all the strips must lie in their original direction4 ,of course you could cut each strip into seven separate pieces1 each piece containing a number1 and the puzzle would then be very easy1 but ,i need hardly say that forty-nine pieces is a long way from being the fewest possible4 #dja4--;,,eight ,,jolly ,,gaol ,,birds4 ,illustration ,the illustration shows the plan of a prison of nine cells all communicating with one another by doorways4 ,the eight prisoners have their numbers on their backs1 and any one of them is allowed to exercise himself in whichever cell may happen to be vacant1 subject to the rule that at no time shall two prisoners be in the same cell4 ,the merry monarch in whose dominions the prison was situated offered them special comforts one ,christmas ,eve if1 without breaking that rule1 they could so place themselves that their numbers should form a magic square4 ,now1 prisoner ,no4 #g happened to know a good deal about magic squares1 so he worked out a scheme and naturally selected the method that was most expeditious--that is1 one involving the fewest possible moves from cell to cell4 ,but one man was a surly1 obstinate fellow 7quite unfit for the society of his jovial companions71 and he refused to move out of his cell or take any part in the proceedings4 ,but ,no4 #g was quite equal to the emergency1 and found that he could still do what was required in the fewest possible moves without troubling the brute to leave his cell4 ,the puzzle is to show how he did it and1 incidentally1 to discover which prisoner was so stupidly obstinate4 ,can you find the fellow8 #djb4--,,nine ,,jolly ,,gaol ,,birds4 ,illustration ,shortly after the episode recorded in the last puzzle occurred1 a ninth prisoner was placed in the vacant cell1 and the merry monarch then offered them all complete liberty on the following strange conditions4 ,they were required so to rearrange themselves in the cells that their numbers formed a magic square without their movements causing any two of them ever to be in the same cell together1 except that at the start one man was allowed to be placed on the shoulders of another man1 and thus add their numbers together1 and move as one man4 ,for example1 ,no4 #h might be placed on the shoulders of ,no4 #b1 and then they would move about together as #aj4 ,the reader should seek first to solve the puzzle in the fewest possible moves1 and then see that the man who is burdened has the least possible amount of work to do4 #djc4--,,the ,,spanish ,,dungeon4 ,not fifty miles from ,cadiz stood in the middle ages a castle1 all traces of which have for centuries disappeared4 ,among other interesting features1 this castle contained a particularly unpleasant dungeon divided into sixteen cells1 all communicating with one another1 as shown in the illustration4 ,now1 the governor was a merry wight1 and very fond of puzzles withal4 ,one day he went to the dungeon and said to the prisoners1 ,by my halidame6 7or its equivalent in ,spanish7 you shall all be set free if you can solve this puzzle4 ,you must so arrange yourselves in the sixteen cells that the numbers on your backs shall form a magic square in which every column1 every row1 and each of the two diagonals shall add up the same4 ,only remember this3 that in no case may two of you ever be together in the same cell4 ,one of the prisoners1 after working at the problem for two or three days1 with a piece of chalk1 undertook to obtain the liberty of himself and his fellow-prisoners if they would follow his directions and move through the doorway from cell to cell in the order in which he should call out their numbers4 ,illustration ,he succeeded in his attempt1 and1 what is more remarkable1 it would seem from the account of his method recorded in the ancient manuscript lying before me1 that he did so in the fewest possible moves4 ,the reader is asked to show what these moves were4 #djd4--,,the ,,siberian ,,dungeons4 ,illustration ,the above is a trustworthy plan of a certain ,russian prison in ,siberia4 ,all the cells are numbered1 and the prisoners are numbered the same as the cells they occupy4 ,the prison diet is so fattening that these political prisoners are in perpetual fear lest1 should their pardon arrive1 they might not be able to squeeze themselves through the narrow doorways and get out4 ,and of course it would be an unreasonable thing to ask any government to pull down the walls of a prison just to liberate the prisoners1 however innocent they might be4 ,therefore these men take all the healthy exercise they can in order to retard their increasing obesity1 and one of their recreations will serve to furnish us with the following puzzle4 ,show1 in the fewest possible moves1 how the sixteen men may form themselves into a magic square1 so that the numbers on their backs shall add up the same in each of the four columns1 four rows1 and two diagonals without two prisoners having been at any time in the same cell together4 ,i had better say1 for the information of those who have not yet been made acquainted with these places1 that it is a peculiarity of prisons that you are not allowed to go outside their walls4 ,any prisoner may go any distance that is possible in a single move4 #dje4--;,,card ,,magic ,,squares4 ,illustration ,take an ordinary pack of cards and throw out the twelve court cards4 ,now1 with nine of the remainder 7different suits are of no consequence7 form the above magic square4 ,it will be seen that the pips add up fifteen in every row in every column1 and in each of the two long diagonals4 ,the puzzle is with the remaining cards 7without disturbing this arrangement7 to form three more such magic squares1 so that each of the four shall add up to a different sum4 ,there will1 of course1 be four cards in the reduced pack that will not be used4 ,these four may be any that you choose4 ,it is not a difficult puzzle1 but requires just a little thought4 #djf4--,,the ,,eighteen ,,dominoes4 ,the illustration shows eighteen dominoes arranged in the form of a square so that the pips in every one of the six columns1 six rows1 and two long diagonals add up #ac4 ,this is the smallest summation possible with any selection of dominoes from an ordinary box of twenty-eight4 ,the greatest possible summation is #bc1 and a solution for this number may be easily obtained by substituting for every number its complement to #f4 ,thus for every blank substitute a #f1 for every #a a #e1 for every #b a #d1 for #c a #c1 for #d a #b1 for #e a #a1 and for #f a blank4 ,but the puzzle is to make a selection of eighteen dominoes and arrange them 7in exactly the form shown7 so that the summations shall be #ah in all the fourteen directions mentioned4 ,illustration ,,subtracting1 ,,multiplying1 ,,and ,,dividing ,,magics4 ,although the adding magic square is of such great antiquity1 curiously enough the multiplying magic does not appear to have been mentioned until the end of the eighteenth century1 when it was referred to slightly by one writer and then forgotten until ,i revived it in ,tit-,bits in #ahig4 ,the dividing magic was apparently first discussed by me in ,the ,weekly ,dispatch in ,june #ahih4 ,the subtracting magic is here introduced for the first time4 ,it will now be convenient to deal with all four kinds of magic squares together4 ,illustration3 ,,adding ,,subtracting ,,multiplying ,,dividing ,in these four diagrams we have examples in the third order of adding1 subtracting1 multiplying1 and dividing squares4 ,in the first the constant1 #ae1 is obtained by the addition of the rows1 columns1 and two diagonals4 ,in the second case you get the constant1 #e1 by subtracting the first number in a line from the second1 and the result from the third4 ,you can1 of course1 perform the operation in either direction2 but1 in order to avoid negative numbers1 it is more convenient simply to deduct the middle number from the sum of the two extreme numbers4 ,this is1 in effect1 the same thing4 ,it will be seen that the constant of the adding square is n times that of the subtracting square derived from it1 where n is the number of cells in the side of square4 ,and the manner of derivation here is simply to reverse the two diagonals4 ,both squares are associated--a term ,i have explained in the introductory article to this department4 ,the third square is a multiplying magic4 ,the constant1 #baf1 is obtained by multiplying together the three numbers in any line4 ,it is associated by multiplication1 instead of by addition4 ,it is here necessary to remark that in an adding square it is not essential that the nine numbers should be consecutive4 ,write down any nine numbers in this way-- #a #c #e #d #f #h #g #i #aa so that the horizontal differences are all alike and the vertical differences also alike 7here #b and #c71 and these numbers will form an adding magic square4 ,by making the differences #a and #c we1 of course1 get consecutive numbers--a particular case1 and nothing more4 ,now1 in the case of the multiplying square we must take these numbers in geometrical instead of arithmetical progression1 thus-- #a #c #i #b #f #ah #d #ab #cf ,here each successive number in the rows is multiplied by #c1 and in the columns by #b4 ,had we multiplied by #b and #h we should get the regular geometrical progression1 #a1 #b1 #d1 #h1 #af1 #cb1 #fd1 #abh1 and #bef1 but ,i wish to avoid high numbers4 ,the numbers are arranged in the square in the same order as in the adding square4 ,the fourth diagram is a dividing magic square4 ,the constant #f is here obtained by dividing the second number in a line by the first 7in either direction7 and the third number by the quotient4 ,but1 again1 the process is simplified by dividing the product of the two extreme numbers by the middle number4 ,this square is also associated by multiplication4 ,it is derived from the multiplying square by merely reversing the diagonals1 and the constant of the multiplying square is the cube of that of the dividing square derived from it4 ,the next set of diagrams shows the solutions for the fifth order of square4 ,they are all associated in the same way as before4 ,the subtracting square is derived from the adding square by reversing the diagonals and exchanging opposite numbers in the centres of the borders1 and the constant of one is again n times that of the other4 ,the dividing square is derived from the multiplying square in the same way1 and the constant of the latter is the #eth power 7that is the nth7 of that of the former4 ,illustration ,these squares are thus quite easy for odd orders4 ,but the reader will probably find some difficulty over the even orders1 concerning which ,i will leave him to make his own researches1 merely propounding two little problems4 #djg4--,,two ,,new ,,magic ,,squares4 ,construct a subtracting magic square with the first sixteen whole numbers that shall be associated by subtraction4 ,the constant is1 of course1 obtained by subtracting the first number from the second in line1 the result from the third1 and the result again from the fourth4 ,also construct a dividing magic square of the same order that shall be associated by division4 ,the constant is obtained by dividing the second number in a line by the first1 the third by the quotient1 and the fourth by the next quotient4 #djh4--,,magic ,,squares ,,of ,,two ,,degrees4 ,while reading a ,french mathematical work ,i happened to come across1 the following statement3 ,a very remarkable magic square of #h1 in two degrees1 has been constructed by ,m4 ,pfeffermann4 ,in other words1 he has managed to dispose the sixty-four first numbers on the squares of a chessboard in such a way that the sum of the numbers in every line1 every column1 and in each of the two diagonals1 shall be the same2 and more1 that if one substitutes for all the numbers their squares1 the square still remains magic4 ,i at once set to work to solve this problem1 and1 although it proved a very hard nut1 one was rewarded by the discovery of some curious and beautiful laws that govern it4 ,the reader may like to try his hand at the puzzle4 ,,magic ,,squares ,,of ,,primes4 ,the problem of constructing magic squares with prime numbers only was first discussed by myself in ,the ,weekly ,dispatch for #bbnd ,july and #eth ,august #aijj2 but during the last three or four years it has received great attention from ,american mathematicians4 ,first1 they have sought to form these squares with the lowest possible constants4 ,thus1 the first nine prime numbers1 #a to #bc inclusive1 sum to #ii1 which 7being divisible by #c7 is theoretically a suitable series2 yet it has been demonstrated that the lowest possible constant is #aaa1 and the required series as follows3 #a1 #g1 #ac1 #ca1 #cg1 #dc1 #fa1 #fg1 and #gc4 ,similarly1 in the case of the fourth order1 the lowest series of primes that are theoretically suitable will not serve4 ,but in every other order1 up to the #abth inclusive1 magic squares have been constructed with the lowest series of primes theoretically possible4 ,and the #abth is the lowest order in which a straight series of prime numbers1 unbroken1 from #a upwards has been made to work4 ,in other words1 the first #add odd prime numbers have actually been arranged in magic form4 ,the following summary is taken from ,the ,monist 7,chicago7 for ,october #aiac3-- ,order of ,totals of ,lowest ,squares ,square4 ,series4 ,constants4 made by-- 7,henry ,e4 #crd #ccc #aaa ,dudeney 7 7#aijj74 7,ernest ,bergholt #dth #djh #ajb and ,c4 ,d4 7 ,shuldham4 #eth #ajfe #bac ,h4 ,a4 ,sayles4 7,c4 ,d4 ,shuldham #fth #bddh #djh and ,j4 7 ,n4 ,muncey4 #gth #dhic #fii do4 #hth #hiab #aaad do4 #ith #aeabi #afha do4 #ajth #bdafj #bdaf ,j4 ,n4 ,muncey4 #aath #cfjie #ccee do4 #abth #edafh #dead do4 ,for further details the reader should consult the article itself1 by ,w4 ,s4 ,andrews and ,h4 ,a4 ,sayles4 ,these same investigators have also performed notable feats in constructing associated and bordered prime magics1 and ,mr4 ,shuldham has sent me a remarkable paper in which he gives examples of ,nasik squares constructed with primes for all orders from the #dth to the #ajth1 with the exception of the #crd 7which is clearly impossible7 and the #ith1 which1 up to the time of writing1 has baffled all attempts4 #dji4--,,the ,,baskets ,,of ,,plums4 ,illustration ,this is the form in which ,i first introduced the question of magic squares with prime numbers4 ,i will here warn the reader that there is a little trap4 ,a fruit merchant had nine baskets4 ,every basket contained plums 7all sound and ripe71 and the number in every basket was different4 ,when placed as shown in the illustration they formed a magic square1 so that if he took any three baskets in a line in the eight possible directions there would always be the same number of plums4 ,this part of the puzzle is easy enough to understand4 ,but what follows seems at first sight a little queer4 ,the merchant told one of his men to distribute the contents of any basket he chose among some children1 giving plums to every child so that each should receive an equal number4 ,but the man found it quite impossible1 no matter which basket he selected and no matter how many children he included in the treat4 ,show1 by giving contents of the nine baskets1 how this could come about4 #daj4--,,the ,,mandarin',s ,t ,,puzzle4 ,illustration ,before ,mr4 ,beauchamp ,cholmondely ,marjoribanks set out on his tour in the ,far ,east1 he prided himself on his knowledge of magic squares1 a subject that he had made his special hobby2 but he soon discovered that he had never really touched more than the fringe of the subject1 and that the wily ,chinee could beat him easily4 ,i present a little problem that one learned mandarin propounded to our traveller1 as depicted on the last page4 ,the ,chinaman1 after remarking that the construction of the ordinary magic square of twenty-five cells is too velly muchee easy1 asked our countryman so to place the numbers #a to #be in the square that every column1 every row1 and each of the two diagonals should add up #fe1 with only prime numbers on the shaded ,t4 ,of course the prime numbers available are #a1 #b1 #c1 #e1 #g1 #aa1 #ac1 #ag1 #ai1 and #bc1 so you are at liberty to select any nine of these that will serve your purpose4 ,can you construct this curious little magic square8 #daa4--;,a ,,magic ,,square ,,of ,,composites4 ,as we have just discussed the construction of magic squares with prime numbers1 the following forms an interesting companion problem4 ,make a magic square with nine consecutive composite numbers--the smallest possible4 #dab4--,,the ,,magic ,,knight',s ,,tour4 ,here is a problem that has never yet been solved1 nor has its impossibility been demonstrated4 ,play the knight once to every square of the chessboard in a complete tour1 numbering the squares in the order visited1 so that when completed the square shall be magic1 adding up to #bfj in every column1 every row1 and each of the two long diagonals4 ,i shall give the best answer that ,i have been able to obtain1 in which there is a slight error in the diagonals alone4 ,can a perfect solution be found8 ,i am convinced that it cannot1 but it is only a pious opinion4 ,,mazes ,,and ,,how ,,to ,,thread ,,them4 ,in wandering mazes lost4 ,paradise ,lost4 ,the ,old ,english word maze1 signifying a labyrinth1 probably comes from the ,scandinavian1 but its origin is somewhat uncertain4 ,the late ,professor ,skeat thought that the substantive was derived from the verb1 and as in old times to be mazed or amazed was to be lost in thought1 the transition to a maze in whose tortuous windings we are lost is natural and easy4 ,the word labyrinth is derived from a ,greek word signifying the passages of a mine4 ,the ancient mines of ,greece and elsewhere inspired fear and awe on account of their darkness and the danger of getting lost in their intricate passages4 ,legend was afterwards built round these mazes4 ,the most familiar instance is the labyrinth made by ,ddalus in ,crete for ,king ,minos4 ,in the centre was placed the ,minotaur1 and no one who entered could find his way out again1 but became the prey of the monster4 ,seven youths and seven maidens were sent regularly by the ,athenians1 and were duly devoured1 until ,theseus slew the monster and escaped from the maze by aid of the clue of thread provided by ,ariadne2 which accounts for our using to-day the expression threading a maze4 ,the various forms of construction of mazes include complicated ranges of caverns1 architectural labyrinths1 or sepulchral buildings1 tortuous devices indicated by coloured marbles and tiled pavements1 winding paths cut in the turf1 and topiary mazes formed by clipped hedges4 ,as a matter of fact1 they may be said to have descended to us in precisely this order of variety4 ,mazes were used as ornaments on the state robes of ,christian emperors before the ninth century1 and were soon adopted in the decoration of cathedrals and other churches4 ,the original idea was doubtless to employ them as symbols of the complicated folds of sin by which man is surrounded4 ,they began to abound in the early part of the twelfth century1 and ,i give an illustration of one of this period in the parish church at ,st4 ,quentin 7,fig4 #a74 ,it formed a pavement of the nave1 and its diameter is #cd feet4 ,the path here is the line itself4 ,if you place your pencil at the point ,a and ignore the enclosing line1 the line leads you to the centre by a long route over the entire area2 but you never have any option as to direction during your course4 ,as we shall find in similar cases1 these early ecclesiastical mazes were generally not of a puzzle nature1 but simply long1 winding paths that took you over practically all the ground enclosed4 ,illustration3 ,,fig4 #a4--,maze at ,st4 ,quentin4 ,illustration3 ,,fig4 #b4--,maze in ,chartres ,cathedral4 ,in the abbey church of ,st4 ,berlin1 at ,st4 ,omer1 is another of these curious floors1 representing the ,temple of ,jerusalem1 with stations for pilgrims4 ,these mazes were actually visited and traversed by them as a compromise for not going to the ,holy ,land in fulfilment of a vow4 ,they were also used as a means of penance1 the penitent frequently being directed to go the whole course of the maze on hands and knees4 ,illustration3 ,,fig4 #c4--,maze in ,lucca ,cathedral4 ,the maze in ,chartres ,cathedral1 of which ,i give an illustration 7,fig4 #b71 is #dj feet across1 and was used by penitents following the procession of ,calvary4 ,a labyrinth in ,amiens ,cathedral was octagonal1 similar to that at ,st4 ,quentin1 measuring #db feet across4 ,it bore the date #abhh1 but was destroyed in #agjh4 ,in the chapter-house at ,bayeux is a labyrinth formed of tiles1 red1 black1 and encaustic1 with a pattern of brown and yellow4 ,dr4 ,ducarel1 in his ,tour through ,part of ,normandy 7printed in #agfg71 mentions the floor of the great guard-chamber in the abbey of ,st4 ,stephen1 at ,caen1 the middle whereof represents a maze or labyrinth about #aj feet diameter1 and so artfully contrived that1 were we to suppose a man following all the intricate meanders of its volutes1 he could not travel less than a mile before he got from one end to the other4 ,illustration3 ,,fig4 #d4--,maze at ,saffron ,walden1 ,essex4 ,then these mazes were sometimes reduced in size and represented on a single tile 7,fig4 #c74 ,i give an example from ,lucca ,cathedral4 ,it is on one of the porch piers1 and is #ai inches in diameter4 ,a writer in #aheh says that1 from the continual attrition it has received from thousands of tracing fingers1 a central group of ,theseus and the ,minotaur has now been very nearly effaced4 ,other examples were1 and perhaps still are1 to be found in the ,abbey of ,toussarts1 at ,chlons-sur-,marne1 in the very ancient church of ,st4 ,michele at ,pavia1 at ,aix in ,provence1 in the cathedrals of ,poitiers1 ,rheims1 and ,arras1 in the church of ,santa ,maria in ,aquiro in ,rome1 in ,san ,vitale at ,ravenna1 in the ,roman mosaic pavement found at ,salzburg1 and elsewhere4 ,these mazes were sometimes called ,chemins de ,jerusalem1 as being emblematical of the difficulties attending a journey to the earthly ,jerusalem and of those encountered by the ,christian before he can reach the heavenly ,jerusalem--where the centre was frequently called ,ciel4 ,common as these mazes were upon the ,continent1 it is probable that no example is to be found in any ,english church2 at least ,i am not aware of the existence of any4 ,but almost every county has1 or has had1 its specimens of mazes cut in the turf4 ,though these are frequently known as miz-mazes or mize-mazes1 it is not uncommon to find them locally called ,troy-towns1 shepherds' races1 or ,julian's ,bowers--names that are misleading1 as suggesting a false origin4 ,from the facts alone that many of these ,english turf mazes are clearly copied from those in the ,continental churches1 and practically all are found close to some ecclesiastical building or near the site of an ancient one1 we may regard it as certain that they were of church origin and not invented by the shepherds or other rustics4 ,and curiously enough1 these turf mazes are apparently unknown on the ,continent4 ,they are distinctly mentioned by ,shakespeare3-- ,the nine men's morris is filled up with mud1 ,and the quaint mazes in the wanton green ,for lack of tread are undistinguishable4 ,a ,midsummer ,night's ,dream1 ii4 #a4 ,my old bones ache3 here's a maze trod indeed1 ,through forth-rights and meanders6 ,the ,tempest1 iii4 #c4 ,illustration3 ,,fig4 #e4--,maze at ,sneinton1 ,nottinghamshire4 ,there was such a maze at ,comberton1 in ,cambridgeshire1 and another1 locally called the miz-maze1 at ,leigh1 in ,dorset4 ,the latter was on the highest part of a field on the top of a hill1 a quarter of a mile from the village1 and was slightly hollow in the middle and enclosed by a bank about #c feet high4 ,it was circular1 and was thirty paces in diameter4 ,in #ahfh the turf had grown over the little trenches1 and it was then impossible to trace the paths of the maze4 ,the ,comberton one was at the same date believed to be perfect1 but whether either or both have now disappeared ,i cannot say4 ,nor have ,i been able to verify the existence or non-existence of the other examples of which ,i am able to give illustrations4 ,i shall therefore write of them all in the past tense1 retaining the hope that some are still preserved4 ,illustration3 ,,fig4 #f4--,maze at ,alkborough1 ,lincolnshire4 ,in the next two mazes given--that at ,saffron ,walden1 ,essex 7#aaj feet in diameter1 ,fig4 #d71 and the one near ,st4 ,anne's ,well1 at ,sneinton1 ,nottinghamshire 7,fig4 #e71 which was ploughed up on ,february #bgth1 #agig 7#ea feet in diameter1 with a path #ece yards long7--the paths must in each case be understood to be on the lines1 black or white1 as the case may be4 ,illustration3 ,,fig4 #g4--,maze at ,boughton ,green1 ,nottinghamshire4 ,i give in ,fig4 #f a maze that was at ,alkborough1 ,lincolnshire1 overlooking the ,humber4 ,this was #dd feet in diameter1 and the resemblance between it and the mazes at ,chartres and ,lucca 7,figs4 #b and #c7 will be at once perceived4 ,a maze at ,boughton ,green1 in ,nottinghamshire1 a place celebrated at one time for its fair 7,fig4 #g71 was #cg feet in diameter4 ,i also include the plan 7,fig4 #h7 of one that used to be on the outskirts of the village of ,wing1 near ,uppingham1 ,rutlandshire4 ,this maze was #dj feet in diameter4 ,illustration3 ,,fig4 #h4--,maze at ,wing1 ,rutlandshire4 ,illustration3 ,,fig4 #i4--,maze on ,st4 ,catherine's ,hill1 ,winchester4 ,the maze that was on ,st4 ,catherine's ,hill1 ,winchester1 in the parish of ,chilcombe1 was a poor specimen 7,fig4 #i71 since1 as will be seen1 there was one short direct route to the centre1 unless1 as in ,fig4 #aj again1 the path is the line itself from end to end4 ,this maze was #hf feet square1 cut in the turf1 and was locally known as the ,mize-maze4 ,it became very indistinct about #aheh1 and was then recut by the ,warden of ,winchester1 with the aid of a plan possessed by a lady living in the neighbourhood4 ,illustration3 ,,fig4 #aj4--,maze on ,ripon ,common4 ,a maze formerly existed on ,ripon ,common1 in ,yorkshire 7,fig4 #aj74 ,it was ploughed up in #ahbg1 but its plan was fortunately preserved4 ,this example was #bj yards in diameter1 and its path is said to have been #djg yards long4 ,illustration3 ,,fig4 #aa4--,maze at ,theobalds1 ,hertfordshire4 ,in the case of the maze at ,theobalds1 ,hertfordshire1 after you have found the entrance within the four enclosing hedges1 the path is forced 7,fig4 #aa74 ,as further illustrations of this class of maze1 ,i give one taken from an ,italian work on architecture by ,serlio1 published in #aecg 7,fig4 #ab71 and one by ,london and ,wise1 the designers of the ,hampton ,court maze1 from their book1 ,the ,retired ,gard'ner1 published in #agjf 7,fig4 #ac74 ,also1 ,i add a ,dutch maze 7,fig4 #ad74 ,illustration3 ,,fig4 #ab4--;,italian ,maze of ,sixteenth ,century4 ,illustration3 ,,fig4 #ac4--;,by the ,designers of ,hampton ,court ,maze4 ,illustration3 ,,fig4 #ad4--;,a ,dutch ,maze4 ,so far our mazes have been of historical interest1 but they have presented no difficulty in threading4 ,after the ,reformation period we find mazes converted into mediums for recreation1 and they generally consisted of labyrinthine paths enclosed by thick and carefully trimmed hedges4 ,these topiary hedges were known to the ,romans1 with whom the topiarius was the ornamental gardener4 ,this type of maze has of late years degenerated into the seaside ,puzzle ,gardens4 ,teas1 sixpence1 including admission to the ,maze4 ,the ,hampton ,court ,maze1 sometimes called the ,wilderness1 at the royal palace1 was designed1 as ,i have said1 by ,london and ,wise for ,william ,,iii41 who had a liking for such things 7,fig4 #ae74 ,i have before me some three or four versions of it1 all slightly different from one another2 but the plan ,i select is taken from an old guide-book to the palace1 and therefore ought to be trustworthy4 ,the meaning of the dotted lines1 etc41 will be explained later on4 ,illustration3 ,,fig4 #ae4--,maze at ,hampton ,court ,palace4 ,illustration3 ,,fig4 #af4--,maze at ,hatfield ,house1 ,herts4 ,illustration3 ,,fig4 #ag4--,maze formerly at ,south ,kensington4 ,illustration3 ,,fig4 #ah4--;,a ,german ,maze4 ,the maze at ,hatfield ,house 7,fig4 #af71 the seat of the ,marquis of ,salisbury1 like so many labyrinths1 is not difficult on paper2 but both this and the ,hampton ,court ,maze may prove very puzzling to actually thread without knowing the plan4 ,one reason is that one is so apt to go down the same blind alleys over and over again1 if one proceeds without method4 ,the maze planned by the desire of the ,prince ,consort for the ,royal ,horticultural ,society's ,gardens at ,south ,kensington was allowed to go to ruin1 and was then destroyed--no great loss1 for it was a feeble thing4 ,it will be seen that there were three entrances from the outside 7,fig4 #ag71 but the way to the centre is very easy to discover4 ,i include a ,german maze that is curious1 but not difficult to thread on paper 7,fig4 #ah74 ,the example of a labyrinth formerly existing at ,pimperne1 in ,dorset1 is in a class by itself 7,fig4 #ai74 ,it was formed of small ridges about a foot high1 and covered nearly an acre of ground2 but it was1 unfortunately1 ploughed up in #agcj4 ,illustration3 ,,fig4 #ai4--,maze at ,pimperne1 ,dorset4 ,we will now pass to the interesting subject of how to thread any maze4 ,while being necessarily brief1 ,i will try to make the matter clear to readers who have no knowledge of mathematics4 ,and first of all we will assume that we are trying to enter a maze 7that is1 get to the centre7 of which we have no plan and about which we know nothing4 ,the first rule is this3 ,if a maze has no parts of its hedges detached from the rest1 then if we always keep in touch with the hedge with the right hand 7or always touch it with the left71 going down to the stop in every blind alley and coming back on the other side1 we shall pass through every part of the maze and make our exit where we went in4 ,therefore we must at one time or another enter the centre1 and every alley will be traversed twice4 ,illustration3 ,,fig4 #bj4--,m4 ,tremaux's ,method of ,solution4 ,illustration3 ,,fig4 #ba4--;,how to thread the ,hatfield ,maze4 ,now look at the ,hampton ,court plan4 ,follow1 say to the right1 the path indicated by the dotted line1 and what ,i have said is clearly correct if we obliterate the two detached parts1 or islands1 situated on each side of the star4 ,but as these islands are there1 you cannot by this method traverse every part of the maze2 and if it had been so planned that the centre was1 like the star1 between the two islands1 you would never pass through the centre at all4 ,a glance at the ,hatfield maze will show that there are three of these detached hedges or islands at the centre1 so this method will never take you to the centre of that one4 ,but the rule will at least always bring you safely out again unless you blunder in the following way4 ,suppose1 when you were going in the direction of the arrow in the ,hampton ,court ,maze1 that you could not distinctly see the turning at the bottom1 that you imagined you were in a blind alley and1 to save time1 crossed at once to the opposite hedge1 then you would go round and round that ,u-shaped island with your right hand still always on the hedge--for ever after6 ,illustration3 ,,fig4 #bb4 ,the ,philadelphia ,maze1 and its ,solution4 ,this blunder happened to me a few years ago in a little maze on the isle of ,caldy1 ,south ,wales4 ,i knew the maze was a small one1 but after a very long walk ,i was amazed to find that ,i did not either reach the centre or get out again4 ,so ,i threw a piece of paper on the ground1 and soon came round to it2 from which ,i knew that ,i had blundered over a supposed blind alley and was going round and round an island4 ,crossing to the opposite hedge and using more care1 ,i was quickly at the centre and out again4 ,now1 if ,i had made a similar mistake at ,hampton ,court1 and discovered the error when at the star1 ,i should merely have passed from one island to another6 ,and if ,i had again discovered that ,i was on a detached part1 ,i might with ill luck have recrossed to the first island again6 ,we thus see that this touching the hedge method should always bring us safely out of a maze that we have entered2 it may happen to take us through the centre1 and if we miss the centre we shall know there must be islands4 ,but it has to be done with a little care1 and in no case can we be sure that we have traversed every alley or that there are no detached parts4 ,illustration3 ,,fig4 #bc4--,simplified ,diagram of ,fig4 #bb4 ,if the maze has many islands1 the traversing of the whole of it may be a matter of considerable difficulty4 ,here is a method for solving any maze1 due to ,m4 ,trmaux1 but it necessitates carefully marking in some way your entrances and exits where the galleries fork4 ,i give a diagram of an imaginary maze of a very simple character that will serve our purpose just as well as something more complex 7,fig4 #bj74 ,the circles at the regions where we have a choice of turnings we may call nodes4 ,a new path or node is one that has not been entered before on the route2 an old path or node is one that has already been entered1 #a4 ,no path may be traversed more than twice4 #b4 ,when you come to a new node1 take any path you like4 #c4 ,when by a new path you come to an old node or to the stop of a blind alley1 return by the path you came4 #d4 ,when by an old path you come to an old node1 take a new path if there is one2 if not1 an old path4 ,the route indicated by the dotted line in the diagram is taken in accordance with these simple rules1 and it will be seen that it leads us to the centre1 although the maze consists of four islands4 ,illustration3 ,,fig4 #bd4--;,can you find the ,shortest ,way to ,centre8 ,neither of the methods ,i have given will disclose to us the shortest way to the centre1 nor the number of the different routes4 ,but we can easily settle these points with a plan4 ,let us take the ,hatfield maze 7,fig4 #ba74 ,it will be seen that ,i have suppressed all the blind alleys by the shading4 ,i begin at the stop and work backwards until the path forks4 ,these shaded parts1 therefore1 can never be entered without our having to retrace our steps4 ,then it is very clearly seen that if we enter at ,a we must come out at ,b2 if we enter at ,c we must come out at ,d4 ,then we have merely to determine whether ,a1 ,b1 ,e1 or ,c1 ,d1 ,e1 is the shorter route4 ,as a matter of fact1 it will be found by rough measurement or calculation that the shortest route to the centre is by way of ,c1 ,d1 ,e1 ,f4 ,illustration3 ,,fig4 #be4--,rosamund's ,bower4 ,i will now give three mazes that are simply puzzles on paper1 for1 so far as ,i know1 they have never been constructed in any other way4 ,the first ,i will call the ,philadelphia maze 7,fig4 #bb74 ,fourteen years ago a travelling salesman1 living in ,philadelphia1 ,u4,s4,a41 developed a curiously unrestrained passion for puzzles4 ,he neglected his business1 and soon his position was taken from him4 ,his days and nights were now passed with the subject that fascinated him1 and this little maze seems to have driven him into insanity4 ,he had been puzzling over it for some time1 and finally it sent him mad and caused him to fire a bullet through his brain4 ,goodness knows what his difficulties could have been6 ,but there can be little doubt that he had a disordered mind1 and that if this little puzzle had not caused him to lose his mental balance some other more or less trivial thing would in time have done so4 ,there is no moral in the story1 unless it be that of the ,irish maxim1 which applies to every occupation of life as much as to the solving of puzzles3 ,take things aisy2 if you can't take them aisy1 take them as aisy as you can4 ,and it is a bad and empirical way of solving any puzzle--by blowing your brains out4 ,now1 how many different routes are there from ,a to ,b in this maze if we must never in any route go along the same passage twice8 ,the four open spaces where four passages end are not reckoned as passages4 ,in the diagram 7,fig4 #bb7 it will be seen that ,i have again suppressed the blind alleys4 ,it will be found that1 in any case1 we must go from ,a to ,c1 and also from ,f to ,b4 ,but when we have arrived at ,c there are three ways1 marked #a1 #b1 #c1 of getting to ,d4 ,similarly1 when we get to ,e there are three ways1 marked #d1 #e1 #f1 of getting to ,f4 ,we have also the dotted route from ,c to ,e1 the other dotted route from ,d to ,f1 and the passage from ,d to ,e1 indicated by stars4 ,we can1 therefore1 express the position of affairs by the little diagram annexed 7,fig4 #bc74 ,here every condition of route exactly corresponds to that in the circular maze1 only it is much less confusing to the eye4 ,now1 the number of routes1 under the conditions1 from ,a to ,b on this simplified diagram is #fdj1 and that is the required answer to the maze puzzle4 ,finally1 ,i will leave two easy maze puzzles 7,figs4 #bd1 #be7 for my readers to solve for themselves4 ,the puzzle in each case is to find the shortest possible route to the centre4 ,everybody knows the story of ,fair ,rosamund and the ,woodstock maze4 ,what the maze was like or whether it ever existed except in imagination is not known1 many writers believing that it was simply a badly-constructed house with a large number of confusing rooms and passages4 ,at any rate1 my sketch lacks the authority of the other mazes in this article4 ,my ,rosamund's ,bower is simply designed to show that where you have the plan before you it often happens that the easiest way to find a route into a maze is by working backwards and first finding a way out4 ,,the ,,paradox ,,party4 ,is not life itself a paradox8 ,c4,l4 ,,dodgson1 ,pillow ,problems4 ,it is a wonderful age6 said ,mr4 ,allgood1 and everybody at the table turned towards him and assumed an attitude of expectancy4 ,this was an ordinary ,christmas dinner of the ,allgood family1 with a sprinkling of local friends4 ,nobody would have supposed that the above remark would lead1 as it did1 to a succession of curious puzzles and paradoxes1 to which every member of the party contributed something of interest4 ,the little symposium was quite unpremeditated1 so we must not be too critical respecting a few of the posers that were forthcoming4 ,the varied character of the contributions is just what we would expect on such an occasion1 for it was a gathering not of expert mathematicians and logicians1 but of quite ordinary folk4 ,it is a wonderful age6 repeated ,mr4 ,allgood4 ,a man has just designed a square house in such a cunning manner that all the windows on the four sides have a south aspect4 ,that would appeal to me1 said ,mrs4 ,allgood1 for ,i cannot endure a room with a north aspect4 ,i cannot conceive how it is done1 ,uncle ,john confessed4 ,i suppose he puts bay windows on the east and west sides2 but how on earth can be contrive to look south from the north side8 ,does he use mirrors1 or something of that kind8 ,no1 replied ,mr4 ,allgood1 nothing of the sort4 ,all the windows are flush with the walls1 and yet you get a southerly prospect from every one of them4 ,you see1 there is no real difficulty in designing the house if you select the proper spot for its erection4 ,now1 this house is designed for a gentleman who proposes to build it exactly at the ,north ,pole4 ,if you think a moment you will realize that when you stand at the ,north ,pole it is impossible1 no matter which way you may turn1 to look elsewhere than due south6 ,there are no such directions as north1 east1 or west when you are exactly at the ,north ,pole4 ,everything is due south6 ,i am afraid1 mother1 said her son ,george1 after the laughter had subsided1 that1 however much you might like the aspect1 the situation would be a little too bracing for you4 ,ah1 well6 she replied4 ,your ,uncle ,john fell also into the trap4 ,i am no good at catches and puzzles4 ,i suppose ,i haven't the right sort of brain4 ,perhaps some one will explain this to me4 ,only last week ,i remarked to my hairdresser that it had been said that there are more persons in the world than any one of them has hairs on his head4 ,he replied1 ',then it follows1 madam1 that two persons1 at least1 must have exactly the same number of hairs on their heads4' ,if this is a fact1 ,i confess ,i cannot see it4 ,how do the bald-headed affect the question8 asked ,uncle ,john4 ,if there are such persons in existence1 replied ,mrs4 ,allgood1 who haven't a solitary hair on their heads discoverable under a magnifying-glass1 we will leave them out of the question4 ,still1 ,i don't see how you are to prove that at least two persons have exactly the same number to a hair4 ,i think ,i can make it clear1 said ,mr4 ,filkins1 who had dropped in for the evening4 ,assume the population of the world to be only one million4 ,any number will do as well as another4 ,then your statement was to the effect that no person has more than nine hundred and ninety-nine thousand nine hundred and ninety-nine hairs on his head4 ,is that so8 ,let me think1 said ,mrs4 ,allgood4 ,yes--yes--that is correct4 ,very well1 then4 ,as there are only nine hundred and ninety-nine thousand nine hundred and ninety-nine different ways of bearing hair1 it is clear that the millionth person must repeat one of those ways4 ,do you see8 ,yes2 ,i see that--at least ,i think ,i see it4 ,therefore two persons at least must have the same number of hairs on their heads2 and as the number of people on the earth so greatly exceeds the number of hairs on any one person's head1 there must1 of course1 be an immense number of these repetitions4 ,but1 ,mr4 ,filkins1 said little ,willie ,allgood1 why could not the millionth man have1 say1 ten thousand hairs and a half8 ,that is mere hair-splitting1 ,willie1 and does not come into the question4 ,here is a curious paradox1 said ,george4 ,if a thousand soldiers are drawn up in battle array on a plane--they understood him to mean plain--only one man will stand upright4 ,nobody could see why4 ,but ,george explained that1 according to ,euclid1 a plane can touch a sphere only at one point1 and that person only who stands at that point1 with respect to the centre of the earth1 will stand upright4 ,in the same way1 he remarked1 if a billiard-table were quite level--that is1 a perfect plane--the balls ought to roll to the centre4 ,though he tried to explain this by placing a visiting-card on an orange and expounding the law of gravitation1 ,mrs4 ,allgood declined to accept the statement4 ,she could not see that the top of a true billiard-table must1 theoretically1 be spherical1 just like a portion of the orange-peel that ,george cut out4 ,of course1 the table is so small in proportion to the surface of the earth that the curvature is not appreciable1 but it is nevertheless true in theory4 ,a surface that we call level is not the same as our idea of a true geometrical plane4 ,uncle ,john1 broke in ,willie ,allgood1 there is a certain island situated between ,england and ,france1 and yet that island is farther from ,france than ,england is4 ,what is the island8 ,that seems absurd1 my boy2 because if ,i place this tumbler1 to represent the island1 between these two plates1 it seems impossible that the tumbler can be farther from either of the plates than they are from each other4 ,but isn't ,guernsey between ,england and ,france8 asked ,willie4 ,yes1 certainly4 ,well1 then1 ,i think you will find1 uncle1 that ,guernsey is about twenty-six miles from ,france1 and ,england is only twenty-one miles from ,france1 between ,calais and ,dover4 ,my mathematical master1 said ,george1 has been trying to induce me to accept the axiom that 'if equals be multiplied by equals the products are equal4' ,it is self-evident1 pointed out ,mr4 ,filkins4 ,for example1 if #c feet equal #a yard1 then twice #c feet will equal #b yards4 ,do you see8 ,but1 ,mr4 ,filkins1 asked ,george1 is this tumbler half full of water equal to a similar glass half empty8 ,certainly1 ,george4 ,then it follows from the axiom that a glass full must equal a glass empty4 ,is that correct8 ,no1 clearly not4 ,i never thought of it in that light4 ,perhaps1 suggested ,mr4 ,allgood1 the rule does not apply to liquids4 ,just what ,i was thinking1 ,allgood4 ,it would seem that we must make an exception in the case of liquids4 ,but it would be awkward1 said ,george1 with a smile1 if we also had to except the case of solids4 ,for instance1 let us take the solid earth4 ,one mile square equals one square mile4 ,therefore two miles square must equal two square miles4 ,is this so8 ,well1 let me see6 ,no1 of course not1 ,mr4 ,filkins replied1 because two miles square is four square miles4 ,then1 said ,george1 if the axiom is not true in these cases1 when is it true8 ,mr4 ,filkins promised to look into the matter1 and perhaps the reader will also like to give it consideration at leisure4 ,look here1 ,george1 said his cousin ,reginald ,woolley3 by what fractional part does four-fourths exceed three-fourths8 ,by one-fourth6 shouted everybody at once4 ,try another one1 ,george suggested4 ,with pleasure1 when you have answered that one correctly1 was ,reginald's reply4 ,do you mean to say that it isn't one-fourth8 ,certainly ,i do4 ,several members of the company failed to see that the correct answer is one-third1 although ,reginald tried to explain that three of anything1 if increased by one-third1 becomes four4 ,uncle ,john1 how do you pronounce 't-o-o'8 asked ,willie4 ',too1 my boy4 ,and how do you pronounce 't-w-o'8 ,that is also 'too4' ,then how do you pronounce the second day of the week8 ,well1 that ,i should pronounce ',tuesday1' not ',toosday4' ,would you really8 ,i should pronounce it ',monday4' ,if you go on like this1 ,willie1 said ,uncle ,john1 with mock severity1 you will soon be without a friend in the world4 ,can any of you write down quickly in figures 'twelve thousand twelve hundred and twelve pounds'8 asked ,mr4 ,allgood4 ,his eldest daughter1 ,miss ,mildred1 was the only person who happened to have a pencil at hand4 ,it can't be done1 she declared1 after making an attempt on the white table-cloth2 but ,mr4 ,allgood showed her that it should be written1 #ac1bab4 ,now it is my turn1 said ,mildred4 ,i have been waiting to ask you all a question4 ,in the ,massacre of the ,innocents under ,herod1 a number of poor little children were buried in the sand with only their feet sticking out4 ,how might you distinguish the boys from the girls8 ,i suppose1 said ,mrs4 ,allgood1 it is a conundrum--something to do with their poor little 'souls4' ,but after everybody had given it up1 ,mildred reminded the company that only boys were put to death4 ,once upon a time1 began ,george1 ,achilles had a race with a tortoise-- ,stop1 ,george6 interposed ,mr4 ,allgood4 ,we won't have that one4 ,i knew two men in my youth who were once the best of friends1 but they quarrelled over that infernal thing of ,zeno's1 and they never spoke to one another again for the rest of their lives4 ,i draw the line at that1 and the other stupid thing by ,zeno about the flying arrow4 ,i don't believe anybody understands them1 because ,i could never do so myself4 ,oh1 very well1 then1 father4 ,here is another4 ,the ,post-,office people were about to erect a line of telegraph-posts over a high hill from ,turmitville to ,wurzleton2 but as it was found that a railway company was making a deep level cutting in the same direction1 they arranged to put up the posts beside the line4 ,now1 the posts were to be a hundred yards apart1 the length of the road over the hill being five miles1 and the length of the level cutting only four and a half miles4 ,how many posts did they save by erecting them on the level8 ,that is a very simple matter of calculation1 said ,mr4 ,filkins4 ,find how many times one hundred yards will go in five miles1 and how many times in four and a half miles4 ,then deduct one from the other1 and you have the number of posts saved by the shorter route4 ,quite right1 confirmed ,mr4 ,allgood4 ,nothing could be easier4 ,that is just what the ,post-,office people said1 replied ,george1 but it is quite wrong4 ,if you look at this sketch that ,i have just made1 you will see that there is no difference whatever4 ,if the posts are a hundred yards apart1 just the same number will be required on the level as over the surface of the hill4 ,illustration ,surely you must be wrong1 ,george1 said ,mrs4 ,allgood1 for if the posts are a hundred yards apart and it is half a mile farther over the hill1 you have to put up posts on that extra half-mile4 ,look at the diagram1 mother4 ,you will see that the distance from post to post is not the distance from base to base measured along the ground4 ,i am just the same distance from you if ,i stand on this spot on the carpet or stand immediately above it on the chair4 ,but ,mrs4 ,allgood was not convinced4 ,mr4 ,smoothly1 the curate1 at the end of the table1 said at this point that he had a little question to ask4 ,suppose the earth were a perfect sphere with a smooth surface1 and a girdle of steel were placed round the ,equator so that it touched at every point4 ',i'll put a girdle round about the earth in forty minutes1' muttered ,george1 quoting the words of ,puck in ,a ,midsummer ,night's ,dream4 ,now1 if six yards were added to the length of the girdle1 what would then be the distance between the girdle and the earth1 supposing that distance to be equal all round8 ,in such a great length1 said ,mr4 ,allgood1 ,i do not suppose the distance would be worth mentioning4 ,what do you say1 ,george8 asked ,mr4 ,smoothly4 ,well1 without calculating ,i should imagine it would be a very minute fraction of an inch4 ,reginald and ,mr4 ,filkins were of the same opinion4 ,i think it will surprise you all1 said the curate1 to learn that those extra six yards would make the distance from the earth all round the girdle very nearly a yard6 ,very nearly a yard6 everybody exclaimed1 with astonishment2 but ,mr4 ,smoothly was quite correct4 ,the increase is independent of the original length of the girdle1 which may be round the earth or round an orange2 in any case the additional six yards will give a distance of nearly a yard all round4 ,this is apt to surprise the non-mathematical mind4 ,did you hear the story of the extraordinary precocity of ,mrs4 ,perkins's baby that died last week8 asked ,mrs4 ,allgood4 ,it was only three months old1 and lying at the point of death1 when the grief-stricken mother asked the doctor if nothing could save it4 ',absolutely nothing6' said the doctor4 ,then the infant looked up pitifully into its mother's face and said--absolutely nothing6 ,impossible6 insisted ,mildred4 ,and only three months old6 ,there have been extraordinary cases of infantile precocity1 said ,mr4 ,filkins1 the truth of which has often been carefully attested4 ,but are you sure this really happened1 ,mrs4 ,allgood8 ,positive1 replied the lady4 ,but do you really think it astonishing that a child of three months should say absolutely nothing8 ,what would you expect it to say8 ,speaking of death1 said ,mr4 ,smoothly1 solemnly1 ,i knew two men1 father and son1 who died in the same battle during the ,south ,african ,war4 ,they were both named ,andrew ,johnson and buried side by side1 but there was some difficulty in distinguishing them on the headstones4 ,what would you have done8 ,quite simple1 said ,mr4 ,allgood4 ,they should have described one as ',andrew ,johnson1 ,senior1' and the other as ',andrew ,johnson1 ,junior4' ,but ,i forgot to tell you that the father died first4 ,what difference can that make8 ,well1 you see1 they wanted to be absolutely exact1 and that was the difficulty4 ,but ,i don't see any difficulty1 said ,mr4 ,allgood1 nor could anybody else4 ,well1 explained ,mr4 ,smoothly1 it is like this4 ,if the father died first1 the son was then no longer ',junior4' ,is that so8 ,to be strictly exact1 yes4 ,that is just what they wanted--to be strictly exact4 ,now1 if he was no longer ',junior1' then he did not die ',junior4 ,consequently it must be incorrect so to describe him on the headstone4 ,do you see the point8 ,here is a rather curious thing1 said ,mr4 ,filkins1 that ,i have just remembered4 ,a man wrote to me the other day that he had recently discovered two old coins while digging in his garden4 ,one was dated '#ea ,b4,c41' and the other one marked ',george ,i4' ,how do ,i know that he was not writing the truth8 ,perhaps you know the man to be addicted to lying1 said ,reginald4 ,but that would be no proof that he was not telling the truth in this instance4 ,perhaps1 suggested ,mildred1 you know that there were no coins made at those dates4 ,on the contrary1 they were made at both periods4 ,were they silver or copper coins8 asked ,willie4 ,my friend did not state1 and ,i really cannot see1 ,willie1 that it makes any difference4 ,i see it6 shouted ,reginald4 ,the letters ',b4,c4' would never be used on a coin made before the birth of ,christ4 ,they never anticipated the event in that way4 ,the letters were only adopted later to denote dates previous to those which we call ',a4,d4' ,that is very good2 but ,i cannot see why the other statement could not be correct4 ,reginald is quite right1 said ,mr4 ,filkins1 about the first coin4 ,the second one could not exist1 because the first ,george would never be described in his lifetime as ',george ,i4' ,why not8 asked ,mrs4 ,allgood4 ,he was ,george ,i4 ,yes2 but they would not know it until there was a ,george ,,ii4 ,then there was no ,george ,,ii4 until ,george ,,iii4 came to the throne8 ,that does not follow4 ,the second ,george becomes ',george ,,ii4' on account of there having been a ',george ,i4' ,then the first ,george was ',george ,i4' on account of there having been no king of that name before him4 ,don't you see1 mother1 said ,george ,allgood1 we did not call ,queen ,victoria ',victoria ,i42' but if there is ever a ',victoria ,,ii41' then she will be known that way4 ,but there have been several ,georges1 and therefore he was ',george ,i4' ,there haven't been several ,victorias1 so the two cases are not similar4 ,they gave up the attempt to convince ,mrs4 ,allgood1 but the reader will1 of course1 see the point clearly4 ,here is a question1 said ,mildred ,allgood1 that ,i should like some of you to settle for me4 ,i am accustomed to buy from our greengrocer bundles of asparagus1 each #ab inches in circumference4 ,i always put a tape measure round them to make sure ,i am getting the full quantity4 ,the other day the man had no large bundles in stock1 but handed me instead two small ones1 each #f inches in circumference4 ',that is the same thing1' ,i said1 'and1 of course1 the price will be the same2' but he insisted that the two bundles together contained more than the large one1 and charged me a few pence extra4 ,now1 what ,i want to know is1 which of us was correct8 ,would the two small bundles contain the same quantity as the large one8 ,or would they contain more8 ,that is the ancient puzzle1 said ,reginald1 laughing1 of the sack of corn that ,sempronius borrowed from ,caius1 which your greengrocer1 perhaps1 had been reading about somewhere4 ,he caught you beautifully4 ,then they were equal8 ,on the contrary1 you were both wrong1 and you were badly cheated4 ,you only got half the quantity that would have been contained in a large bundle1 and therefore ought to have been charged half the original price1 instead of more4 ,yes1 it was a bad swindle1 undoubtedly4 ,a circle with a circumference half that of another must have its area a quarter that of the other4 ,therefore the two small bundles contained together only half as much asparagus as a large one4 ,mr4 ,filkins1 can you answer this8 asked ,willie4 ,there is a man in the next village who eats two eggs for breakfast every morning4 ,nothing very extraordinary in that1 ,george broke in4 ,if you told us that the two eggs ate the man it would be interesting4 ,don't interrupt the boy1 ,george1 said his mother4 ,well1 ,willie continued1 this man neither buys1 borrows1 barters1 begs1 steals1 nor finds the eggs4 ,he doesn't keep hens1 and the eggs are not given to him4 ,how does he get the eggs8 ,does he take them in exchange for something else8 asked ,mildred4 ,that would be bartering them1 ,willie replied4 ,perhaps some friend sends them to him1 suggested ,mrs4 ,allgood4 ,i said that they were not given to him4 ,i know1 said ,george1 with confidence4 ,a strange hen comes into his place and lays them4 ,but that would be finding them1 wouldn't it8 ,does he hire them8 asked ,reginald4 ,if so1 he could not return them after they were eaten1 so that would be stealing them4 ,perhaps it is a pun on the word 'lay1' ,mr4 ,filkins said4 ,does he lay them on the table8 ,he would have to get them first1 wouldn't he8 ,the question was1 ,how does he get them8 ,give it up6 said everybody4 ,then little ,willie crept round to the protection of his mother1 for ,george was apt to be rough on such occasions4 ,the man keeps ducks6 he cried1 and his servant collects the eggs every morning4 ,but you said he doesn't keep birds6 ,george protested4 ,i didn't1 did ,i1 ,mr4 ,filkins8 ,i said he doesn't keep hens4 ,but he finds them1 said ,reginald4 ,no2 ,i said his servant finds them4 ,well1 then1 ,mildred interposed1 his servant gives them to him4 ,you cannot give a man his own property1 can you8 ,all agreed that ,willie's answer was quite satisfactory4 ,then ,uncle ,john produced a little fallacy that brought the proceedings to a close1 as the newspapers say4 #dac4--;,a ,,chessboard ,,fallacy4 ,illustration ,here is a diagram of a chessboard1 he said4 ,you see there are sixty-four squares--eight by eight4 ,now ,i draw a straight line from the top left-hand corner1 where the first and second squares meet1 to the bottom right-hand corner4 ,i cut along this line with the scissors1 slide up the piece that ,i have marked ,b1 and then clip off the little corner ,c by a cut along the first upright line4 ,this little piece will exactly fit into its place at the top1 and we now have an oblong with seven squares on one side and nine squares on the other4 ,there are1 therefore1 now only sixty-three squares1 because seven multiplied by nine makes sixty-three4 ,where on earth does that lost square go to8 ,i have tried over and over again to catch the little beggar1 but he always eludes me4 ,for the life of me ,i cannot discover where he hides himself4 ,it seems to be like the other old chessboard fallacy1 and perhaps the explanation is the same1 said ,reginald--that the pieces do not exactly fit4 ,but they do fit1 said ,uncle ,john4 ,try it1 and you will see4 ,later in the evening ,reginald and ,george1 were seen in a corner with their heads together1 trying to catch that elusive little square1 and it is only fair to record that before they retired for the night they succeeded in securing their prey1 though some others of the company failed to see it when captured4 ,can the reader solve the little mystery8 ,,unclassified ,,problems4 ,a snapper up of unconsidered trifles4 ,winter's ,tale1 iv4 #b4 #dad4--,,who ,,was ,,first8 ,anderson1 ,biggs1 and ,carpenter were staying together at a place by the seaside4 ,one day they went out in a boat and were a mile at sea when a rifle was fired on shore in their direction4 ,why or by whom the shot was fired fortunately does not concern us1 as no information on these points is obtainable1 but from the facts ,i picked up we can get material for a curious little puzzle for the novice4 ,it seems that ,anderson only heard the report of the gun1 ,biggs only saw the smoke1 and ,carpenter merely saw the bullet strike the water near them4 ,now1 the question arises3 ,which of them first knew of the discharge of the rifle8 #dae4--;,a ,,wonderful ,,village4 ,there is a certain village in ,japan1 situated in a very low valley1 and yet the sun is nearer to the inhabitants every noon1 by #c1jjj miles and upwards1 than when he either rises or sets to these people4 ,in what part of the country is the village situated8 #daf4--;,a ,,calendar ,,puzzle4 ,if the end of the world should come on the first day of a new century1 can you say what are the chances that it will happen on a ,sunday8 #dag4--,,the ,,tiring ,,irons4 ,illustration ,the illustration represents one of the most ancient of all mechanical puzzles4 ,its origin is unknown4 ,cardan1 the mathematician1 wrote about it in #aeej1 and ,wallis in #afic2 while it is said still to be found in obscure ,english villages 7sometimes deposited in strange places1 such as a church belfry71 made of iron1 and appropriately called tiring-irons1 and to be used by the ,norwegians to-day as a lock for boxes and bags4 ,in the toyshops it is sometimes called the ,chinese rings1 though there seems to be no authority for the description1 and it more frequently goes by the unsatisfactory name of the puzzling rings4 ,the ,french call it ,baguenaudier4 ,the puzzle will be seen to consist of a simple loop of wire fixed in a handle to be held in the left hand1 and a certain number of rings secured by wires which pass through holes in the bar and are kept there by their blunted ends4 ,the wires work freely in the bar1 but cannot come apart from it1 nor can the wires be removed from the rings4 ,the general puzzle is to detach the loop completely from all the rings1 and then to put them all on again4 ,now1 it will be seen at a glance that the first ring 7to the right7 can be taken off at any time by sliding it over the end and dropping it through the loop2 or it may be put on by reversing the operation4 ,with this exception1 the only ring that can ever be removed is the one that happens to be a contiguous second on the loop at the right-hand end4 ,thus1 with all the rings on1 the second can be dropped at once2 with the first ring down1 you cannot drop the second1 but may remove the third2 with the first three rings down1 you cannot drop the fourth1 but may remove the fifth2 and so on4 ,it will be found that the first and second rings can be dropped together or put on together2 but to prevent confusion we will throughout disallow this exceptional double move1 and say that only one ring may be put on or removed at a time4 ,we can thus take off one ring in #a move2 two rings in #b moves2 three rings in #e moves2 four rings in #aj moves2 five rings in #ba moves2 and if we keep on doubling 7and adding one where the number of rings is odd7 we may easily ascertain the number of moves for completely removing any number of rings4 ,to get off all the seven rings requires #he moves4 ,let us look at the five moves made in removing the first three rings1 the circles above the line standing for rings on the loop and those under for rings off the loop4 ,drop the first ring2 drop the third2 put up the first2 drop the second2 and drop the first--#e moves1 as shown clearly in the diagrams4 ,the dark circles show at each stage1 from the starting position to the finish1 which rings it is possible to drop4 ,after move #b it will be noticed that no ring can be dropped until one has been put on1 because the first and second rings from the right now on the loop are not together4 ,after the fifth move1 if we wish to remove all seven rings we must now drop the fifth4 ,but before we can then remove the fourth it is necessary to put on the first three and remove the first two4 ,we shall then have #g1 #f1 #d1 #c on the loop1 and may therefore drop the fourth4 ,when we have put on #b and #a and removed #c1 #b1 #a1 we may drop the seventh ring4 ,the next operation then will be to get #f1 #e1 #d1 #c1 #b1 #a on the loop and remove #d1 #c1 #b1 #a1 when #f will come off2 then get #e1 #d1 #c1 #b1 #a on the loop1 and remove #c1 #b1 #a1 when #e will come off2 then get #d1 #c1 #b1 #a on the loop and remove #b1 #a1 when #d will come off2 then get #c1 #b1 #a on the loop and remove #a1 when #c will come off2 then get #b1 #a on the loop1 when #b will come off2 and #a will fall through on the #heth move1 leaving the loop quite free4 ,the reader should now be able to understand the puzzle1 whether or not he has it in his hand in a practical form4 ,illustration ,illustration3 o o o o o 99 99 ------------- o o o o 99 o #a------------- o o o o o o #b------------- o o o o o o 99 99 #c------------- o o o o o 99 #d------------- o o o o 99 o #e------------- o o o ,the particular problem ,i propose is simply this4 ,suppose there are altogether fourteen rings on the tiring-irons1 and we proceed to take them all off in the correct way so as not to waste any moves4 ,what will be the position of the rings after the #i1iiith move has been made8 #dah4--,,such ,a ,,getting ,,upstairs4 ,in a suburban villa there is a small staircase with eight steps1 not counting the landing4 ,the little puzzle with which ,tommy ,smart perplexed his family is this4 ,you are required to start from the bottom and land twice on the floor above 7stopping there at the finish71 having returned once to the ground floor4 ,but you must be careful to use every tread the same number of times4 ,in how few steps can you make the ascent8 ,it seems a very simple matter1 but it is more than likely that at your first attempt you will make a great many more steps than are necessary4 ,of course you must not go more than one riser at a time4 ,tommy knows the trick1 and has shown it to his father1 who professes to have a contempt for such things2 but when the children are in bed the pater will often take friends out into the hall and enjoy a good laugh at their bewilderment4 ,and yet it is all so very simple when you know how it is done4 #dai4--,,the ,,five ,,pennies4 ,here is a really hard puzzle1 and yet its conditions are so absurdly simple4 ,every reader knows how to place four pennies so that they are equidistant from each other4 ,all you have to do is to arrange three of them flat on the table so that they touch one another in the form of a triangle1 and lay the fourth penny on top in the centre4 ,then1 as every penny touches every other penny1 they are all at equal distances from one another4 ,now try to do the same thing with five pennies--place them so that every penny shall touch every other penny--and you will find it a different matter altogether4 #dbj4--,,the ,,industrious ,,bookworm4 ,illustration ,our friend ,professor ,rackbrane is seen in the illustration to be propounding another of his little posers4 ,he is explaining that since he last had occasion to take down those three volumes of a learned book from their place on his shelves a bookworm has actually bored a hole straight through from the first page to the last4 ,he says that the leaves are together three inches thick in each volume1 and that every cover is exactly one-eighth of an inch thick1 and he asks how long a tunnel had the industrious worm to bore in preparing his new tube railway4 ,can you tell him8 #dba4--;,a ,,chain ,,puzzle4 ,illustration ,this is a puzzle based on a pretty little idea first dealt with by the late ,mr4 ,sam ,loyd4 ,a man had nine pieces of chain1 as shown in the illustration4 ,he wanted to join these fifty links into one endless chain4 ,it will cost a penny to open any link and twopence to weld a link together again1 but he could buy a new endless chain of the same character and quality for #bs4 #b;d4 ,what was the cheapest course for him to adopt8 ,unless the reader is cunning he may find himself a good way out in his answer4 #dbb4--,,the ,,sabbath ,,puzzle4 ,i have come across the following little poser in an old book4 ,i wonder how many readers will see the author's intended solution to the riddle4 ,christians the week's first day for ,sabbath hold2 ,the ,jews the seventh1 as they did of old2 ,the ,turks the sixth1 as we have oft been told4 ,how can these three1 in the same place and day1 ,have each his own true ,sabbath8 tell1 ,i pray4 #dbc4--,,the ,,ruby ,,brooch4 ,the annals of ,scotland ,yard contain some remarkable cases of jewel robberies1 but one of the most perplexing was the theft of ,lady ,littlewood's rubies4 ,there have1 of course1 been many greater robberies in point of value1 but few so artfully conceived4 ,lady ,littlewood1 of ,romley ,manor1 had a beautiful but rather eccentric heirloom in the form of a ruby brooch4 ,while staying at her town house early in the eighties she took the jewel to a shop in ,brompton for some slight repairs4 ,a fine collection of rubies1 madam1 said the shopkeeper1 to whom her ladyship was a stranger4 ,yes1 she replied2 but curiously enough ,i have never actually counted them4 ,my mother once pointed out to me that if you start from the centre and count up one line1 along the outside and down the next line1 there are always eight rubies4 ,so ,i should always know if a stone were missing4 ,illustration ,six months later a brother of ,lady ,littlewood's1 who had returned from his regiment in ,india1 noticed that his sister was wearing the ruby brooch one night at a county ball1 and on their return home asked to look at it more closely4 ,he immediately detected the fact that four of the stones were gone4 ,how can that possibly be8 said ,lady ,littlewood4 ,if you count up one line from the centre1 along the edge1 and down the next line1 in any direction1 there are always eight stones4 ,this was always so and is so now4 ,how1 therefore1 would it be possible to remove a stone without my detecting it8 ,nothing could be simpler1 replied the brother4 ,i know the brooch well4 ,it originally contained forty-five stones1 and there are now only forty-one4 ,somebody has stolen four rubies1 and then reset as small a number of the others as possible in such a way that there shall always be eight in any of the directions you have mentioned4 ,there was not the slightest doubt that the ,brompton jeweller was the thief1 and the matter was placed in the hands of the police4 ,but the man was wanted for other robberies1 and had left the neighbourhood some time before4 ,to this day he has never been found4 ,the interesting little point that at first baffled the police1 and which forms the subject of our puzzle1 is this3 ,how were the forty-five rubies originally arranged on the brooch8 ,the illustration shows exactly how the forty-one were arranged after it came back from the jeweller2 but although they count eight correctly in any of the directions mentioned1 there are four stones missing4 #dbd4--,,the ,,dovetailed ,,block4 ,illustration ,here is a curious mechanical puzzle that was given to me some years ago1 but ,i cannot say who first invented it4 ,it consists of two solid blocks of wood securely dovetailed together4 ,on the other two vertical sides that are not visible the appearance is precisely the same as on those shown4 ,how were the pieces put together8 ,when ,i published this little puzzle in a ,london newspaper ,i received 7though they were unsolicited7 quite a stack of models1 in oak1 in teak1 in mahogany1 rosewood1 satinwood1 elm1 and deal2 some half a foot in length1 and others varying in size right down to a delicate little model about half an inch square4 ,it seemed to create considerable interest4 #dbe4--;,,jack ,,and ,,the ,,beanstalk4 ,illustration ,the illustration1 by a ,british artist1 is a sketch of ,jack climbing the beanstalk4 ,now1 the artist has made a serious blunder in this drawing4 ,can you find out what it is8 #dbf4--,,the ,,hymn-,,board ,,poser4 ,the worthy vicar of ,chumpley ,st4 ,winifred is in great distress4 ,a little church difficulty has arisen that all the combined intelligence of the parish seems unable to surmount4 ,what this difficulty is ,i will state hereafter1 but it may add to the interest of the problem if ,i first give a short account of the curious position that has been brought about4 ,it all has to do with the church hymn-boards1 the plates of which have become so damaged that they have ceased to fulfil the purpose for which they were devised4 ,a generous parishioner has promised to pay for a new set of plates at a certain rate of cost2 but strange as it may seem1 no agreement can be come to as to what that cost should be4 ,the proposed maker of the plates has named a price which the donor declares to be absurd4 ,the good vicar thinks they are both wrong1 so he asks the schoolmaster to work out the little sum4 ,but this individual declares that he can find no rule bearing on the subject in any of his arithmetic books4 ,an application having been made to the local medical practitioner1 as a man of more than average intellect at ,chumpley1 he has assured the vicar that his practice is so heavy that he has not had time even to look at it1 though his assistant whispers that the doctor has been sitting up unusually late for several nights past4 ,widow ,wilson has a smart son1 who is reputed to have once won a prize for puzzle-solving4 ,he asserts that as he cannot find any solution to the problem it must have something to do with the squaring of the circle1 the duplication of the cube1 or the trisection of an angle2 at any rate1 he has never before seen a puzzle on the principle1 and he gives it up4 ,illustration ,this was the state of affairs when the assistant curate 7who1 ,i should say1 had frankly confessed from the first that a profound study of theology had knocked out of his head all the knowledge of mathematics he ever possessed7 kindly sent me the puzzle4 ,a church has three hymn-boards1 each to indicate the numbers of five different hymns to be sung at a service4 ,all the boards are in use at the same service4 ,the hymn-book contains #gjj hymns4 ,a new set of numbers is required1 and a kind parishioner offers to present a set painted on metal plates1 but stipulates that only the smallest number of plates necessary shall be purchased4 ,the cost of each plate is to be #f;d41 and for the painting of each plate the charges are to be3 ,for one plate1 #as42 for two plates alike1 #aa;d4 each2 for three plates alike1 #aa;d4 each1 and so on1 the charge being one farthing less per plate for each similarly painted plate4 ,now1 what should be the lowest cost8 ,readers will note that they are required to use every legitimate and practical method of economy4 ,the illustration will make clear the nature of the three hymn-boards and plates4 ,the five hymns are here indicated by means of twelve plates4 ,these plates slide in separately at the back1 and in the illustration there is room1 of course1 for three more plates4 #dbg4--,,pheasant-,,shooting4 ,a ,cockney friend1 who is very apt to draw the long bow1 and is evidently less of a sportsman than he pretends to be1 relates to me the following not very credible yarn3-- ,i've just been pheasant-shooting with my friend the duke4 ,we had splendid sport1 and ,i made some wonderful shots4 ,what do you think of this1 for instance8 ,perhaps you can twist it into a puzzle4 ,the duke and ,i were crossing a field when suddenly twenty-four pheasants rose on the wing right in front of us4 ,i fired1 and two-thirds of them dropped dead at my feet4 ,then the duke had a shot at what were left1 and brought down three-twenty-fourths of them1 wounded in the wing4 ,now1 out of those twenty-four birds1 how many still remained8 ,it seems a simple enough question1 but can the reader give a correct answer8 #dbh4--,,the ,,gardener ,,and ,,the ,,cook4 ,a correspondent1 signing himself ,simple ,simon1 suggested that ,i should give a special catch puzzle in the issue of ,the ,weekly ,dispatch for ,all ,fools' ,day1 #aijj4 ,so ,i gave the following1 and it caused considerable amusement2 for out of a very large body of competitors1 many quite expert1 not a single person solved it1 though it ran for nearly a month4 ,illustration ,the illustration is a fancy sketch of my correspondent1 ',simple ,simon1' in the act of trying to solve the following innocent little arithmetical puzzle4 ,a race between a man and a woman that ,i happened to witness one ,all ,fools' ,day has fixed itself indelibly on my memory4 ,it happened at a country-house1 where the gardener and the cook decided to run a race to a point #ajj feet straight away and return4 ,i found that the gardener ran #c feet at every bound and the cook only #b feet1 but then she made three bounds to his two4 ,now1 what was the result of the race8 ,a fortnight after publication ,i added the following note3 ,it has been suggested that perhaps there is a catch in the 'return1' but there is not4 ,the race is to a point #ajj feet away and home again--that is1 a distance of #bjj feet4 ,one correspondent asks whether they take exactly the same time in turning1 to which ,i reply that they do4 ,another seems to suspect that it is really a conundrum1 and that the answer is that 'the result of the race was a 7matrimonial7 tie4' ,but ,i had no such intention4 ,the puzzle is an arithmetical one1 as it purports to be4 #dbi4--,,placing ,,halfpennies4 ,illustration ,here is an interesting little puzzle suggested to me by ,mr4 ,w4 ,t4 ,whyte4 ,mark off on a sheet of paper a rectangular space #e inches by #c inches1 and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions4 ,a halfpenny is exactly an inch in diameter4 ,place your first halfpenny where you like1 then place your second coin at exactly the distance of an inch from the first1 the third an inch distance from the second1 and so on4 ,no halfpenny may touch another halfpenny or cross the boundary4 ,our illustration will make the matter perfectly clear4 ,no4 #b coin is an inch from ,no4 #a2 ,no4 #c an inch from ,no4 #b2 ,no4 #d an inch from ,no4 #c2 but after ,no4 #aj is placed we can go no further in this attempt4 ,yet several more halfpennies might have been got in4 ,how many can the reader place8 #dcj4--;,,find ,,the ,,man',s ,,wife4 ,illustration ,one summer day in #aijc ,i was loitering on the ,brighton front1 watching the people strolling about on the beach1 when the friend who was with me suddenly drew my attention to an individual who was standing alone1 and said1 ,can you point out that man's wife8 ,they are stopping at the same hotel as ,i am1 and the lady is one of those in view4 ,after a few minutes' observation1 ,i was successful in indicating the lady correctly4 ,my friend was curious to know by what method of reasoning ,i had arrived at the result4 ,this was my answer3-- ,we may at once exclude that ,sister of ,mercy and the girl in the short frock2 also the woman selling oranges4 ,it cannot be the lady in widows' weeds4 ,it is not the lady in the bath chair1 because she is not staying at your hotel1 for ,i happened to see her come out of a private house this morning assisted by her maid4 ,the two ladies in red breakfasted at my hotel this morning1 and as they were not wearing outdoor dress ,i conclude they are staying there4 ,it therefore rests between the lady in blue and the one with the green parasol4 ,but the left hand that holds the parasol is1 you see1 ungloved and bears no wedding-ring4 ,consequently ,i am driven to the conclusion that the lady in blue is the man's wife--and you say this is correct4 ,now1 as my friend was an artist1 and as ,i thought an amusing puzzle might be devised on the lines of his question1 ,i asked him to make me a drawing according to some directions that ,i gave him1 and ,i have pleasure in presenting his production to my readers4 ,it will be seen that the picture shows six men and six ladies3 ,nos4 #a1 #c1 #e1 #g1 #i1 and #aa are ladies1 and ,nos4 #b1 #d1 #f1 #h1 #aj1 and #ab are men4 ,these twelve individuals represent six married couples1 all strangers to one another1 who1 in walking aimlessly about1 have got mixed up4 ,but we are only concerned with the man that is wearing a straw hat--,number #aj4 ,the puzzle is to find this man's wife4 ,examine the six ladies carefully1 and see if you can determine which one of them it is4 ,i showed the picture at the time to a few friends1 and they expressed very different opinions on the matter4 ,one said1 ,i don't believe he would marry a girl like ,number #g4 ,another said1 ,i am sure a nice girl like ,number #c would not marry such a fellow6 ,another said1 ,it must be ,number #a1 because she has got as far away as possible from the brute6 ,it was suggested1 again1 that it must be ,number #aa1 because he seems to be looking towards her2 but a cynic retorted1 ,for that very reason1 if he is really looking at her1 ,i should say that she is not his wife6 ,i now leave the question in the hands of my readers4 ,which is really ,number #aj's wife8 ,the illustration is of necessity considerably reduced from the large scale on which it originally appeared in ,the ,weekly ,dispatch 7#bdth ,may #aijc71 but it is hoped that the details will be sufficiently clear to allow the reader to derive entertainment from its examination4 ,in any case the solution given will enable him to follow the points with interest4 ,,solutions4 #a4--;,a ,,post-,,office ,,perplexity4 ,the young lady supplied #e twopenny stamps1 #cj penny stamps1 and #h twopence-halfpenny stamps1 which delivery exactly fulfils the conditions and represents a cost of five shillings4 #b4--,,youthful ,,precocity4 ,the price of the banana must have been one penny farthing4 ,thus1 #ifj bananas would cost #e1 and #dhj sixpences would buy #b1cjd bananas4 #c4--;,,at ,a ,,cattle ,,market4 ,jakes must have taken #g animals to market1 ,hodge must have taken #aa1 and ,durrant must have taken #ba4 ,there were thus #ci animals altogether4 #d4--,,the ,,beanfeast ,,puzzle4 ,the cobblers spent #ces41 the tailors spent also #ces41 the hatters spent #dbs41 and the glovers spent #bas4 ,thus1 they spent altogether #f1acs41 while it will be found that the five cobblers spent as much as four tailors1 twelve tailors as much as nine hatters1 and six hatters as much as eight glovers4 #e4--;,a ,,queer ,,coincidence4 ,puzzles of this class are generally solved in the old books by the tedious process of working backwards4 ,but a simple general solution is as follows3 ,if there are n players1 the amount held by every player at the end will be m7#bn71 the last winner must have held m7n #a7 at the start1 the next m7#bn #a71 the next m7#dn #a71 the next m7#hn #a71 and so on to the first player1 who must have held m7#bn - #an #a74 ,thus1 in this case1 n #g1 and the amount held by every player at the end was #bg farthings4 ,therefore m #a1 and ,g started with #h farthings1 ,f with #ae1 ,e with #bi1 ,d with #eg1 ,c with #aac1 ,b with #bbe1 and ,a with #ddi farthings4 #f4--;,a ,,charitable ,,bequest4 ,there are seven different ways in which the money may be distributed3 #e women and #ai men1 #aj women and #af men1 #ae women and #ac men1 #bj women and #aj men1 #be women and #g men1 #cj women and #d men1 and #ce women and #a man4 ,but the last case must not be counted1 because the condition was that there should be men1 and a single man is not men4 ,therefore the answer is six years4 #g4--,,the ,,widow',s ,,legacy4 ,the widow's share of the legacy must be #bje1 #bs4 #f;d4 and #aj_/ac of a penny4 #h4--;,,indiscriminate ,,charity ,the gentleman must have had #cs4 #f;d4 in his pocket when he set out for home4 #i4--,,the ,,two ,,aeroplanes4 ,the man must have paid #ejj and #gej for the two machines1 making together #a1bej2 but as he sold them for only #a1bjj1 he lost #ej by the transaction4 #aj4--;,,buying ,,presents4 ,jorkins had originally #ai1 #ahs4 in his pocket1 and spent #i1 #ais4 #aa4--,,the ,,cyclists' ,,feast4 ,there were ten cyclists at the feast4 ,they should have paid #hs4 each2 but1 owing to the departure of two persons1 the remaining eight would pay #ajs4 each4 #ab4--;,a ,,queer ,,thing ,,in ,,money4 ,the answer is as follows3 #dd1ddd1 #ds4 #d;d4 #bh1 and1 reduced to pence1 #aj1fff1fabbh4 ,it is a curious little coincidence that in the answer #aj1fff1fab the four central figures indicate the only other answer1 #ff1 #fs4 #f;d4 #ac4--;,a ,,new ,,money ,,puzzle4 ,the smallest sum of money1 in pounds1 shillings1 pence1 and farthings1 containing all the nine digits once1 and once only1 is #b1efg1 #ahs4 #i;d4 #ad4--,,square ,,money4 ,the answer is #a;d4 and #c;d4 ,added together they make #d;d41 and #a;d4 multiplied by #c is also #d;d4 #ae4--,,pocket ,,money4 ,the largest possible sum is #aes4 #i;d41 composed of a crown and a half-crown 7or three half-crowns71 four florins1 and a threepenny piece4 #af4--,,the ,,millionaire',s ,,perplexity4 ,the answer to this quite easy puzzle may1 of course1 be readily obtained by trial1 deducting the largest power of #g that is contained in one million dollars1 then the next largest power from the remainder1 and so on4 ,but the little problem is intended to illustrate a simple direct method4 ,the answer is given at once by converting #a1jjj1jjj to the septenary scale1 and it is on this subject of scales of notation that ,i propose to write a few words for the benefit of those who have never sufficiently considered the matter4 ,our manner of figuring is a sort of perfected arithmetical shorthand1 a system devised to enable us to manipulate numbers as rapidly and correctly as possible by means of symbols4 ,if we write the number #b1cda to represent two thousand three hundred and forty-one dollars1 we wish to imply #a dollar1 added to four times #aj dollars1 added to three times #ajj dollars1 added to two times #a1jjj dollars4 ,from the number in the units place on the right1 every figure to the left is understood to represent a multiple of the particular power of #aj that its position indicates1 while a cipher 7#j7 must be inserted where necessary in order to prevent confusion1 for if instead of #bjg we wrote #bg it would be obviously misleading4 ,we thus only require ten figures1 because directly a number exceeds #i we put a second figure to the left1 directly it exceeds #ii we put a third figure to the left1 and so on4 ,it will be seen that this is a purely arbitrary method4 ,it is working in the denary 7or ten7 scale of notation1 a system undoubtedly derived from the fact that our forefathers who devised it had ten fingers upon which they were accustomed to count1 like our children of to-day4 ,it is unnecessary for us ordinarily to state that we are using the denary scale1 because this is always understood in the common affairs of life4 ,but if a man said that he had #f1eec dollars in the septenary 7or seven7 scale of notation1 you will find that this is precisely the same amount as #b1cda in our ordinary denary scale4 ,instead of using powers of ten1 he uses powers of #g1 so that he never needs any figure higher than #f1 and #f1eec really stands for #c1 added to five times #g1 added to five times #di1 added to six times #cdc 7in the ordinary notation71 or #b1cda4 ,to reverse the operation1 and convert #b1cda from the denary to the septenary scale1 we divide it by #g1 and get #ccd and remainder #c2 divide #ccd by #g1 and get #dg and remainder #e2 and so keep on dividing by #g as long as there is anything to divide4 ,the remainders1 read backwards1 #f1 #e1 #e1 #c1 give us the answer1 #f1eec4 ,now1 as ,i have said1 our puzzle may be solved at once by merely converting #a1jjj1jjj dollars to the septenary scale4 ,keep on dividing this number by #g until there is nothing more left to divide1 and the remainders will be found to be #aaccccaa which is #a1jjj1jjj expressed in the septenary scale4 ,therefore1 #a gift of #a dollar1 #a gift of #g dollars1 #c gifts of #di dollars1 #c gifts of #cdc dollars1 #c gifts of #b1dja dollars1 #c gifts of #af1hjg dollars1 #a gift of #aag1fdi dollars1 and one substantial gift of #hbc1edc dollars1 satisfactorily solves our problem4 ,and it is the only possible solution4 ,it is thus seen that no trials are necessary2 by converting to the septenary scale of notation we go direct to the answer4 #ag4--,,the ,,puzzling ,,money ,,boxes4 ,the correct answer to this puzzle is as follows3 ,john put into his money-box two double florins 7#hs471 ,william a half-sovereign and a florin 7#abs471 ,charles a crown 7#es471 and ,thomas a sovereign 7#bjs474 ,there are six coins in all1 of a total value of #des4 ,if ,john had #bs4 more1 ,william #bs4 less1 ,charles twice as much1 and ,thomas half as much as they really possessed1 they would each have had exactly #ajs4 #ah4--,,the ,,market ,,women4 ,the price received was in every case #aje farthings4 ,therefore the greatest number of women is eight1 as the goods could only be sold at the following rates3 #aje lbs4 at #a farthing1 #ce at #c1 #ba at #e1 #ae at #g1 #g at #ae1 #e at #ba1 #c at #ce1 and #a lb4 at #aje farthings4 #ai4--,,the ,,new ,,year',s ,,eve ,,suppers4 ,the company present on the occasion must have consisted of seven pairs1 ten single men1 and one single lady4 ,thus1 there were twenty-five persons in all1 and at the prices stated they would pay exactly #e together4 #bj4--;,,beef ,,and ,,sausages4 ,the lady bought #dh lbs4 of beef at #bs41 and the same quantity of sausages at #as4 #f;d41 thus spending #h1 #hs4 ,had she bought #db lbs4 of beef and #ef lbs4 of sausages she would have spent #d1 #ds4 on each1 and have obtained #ih lbs4 instead of #if lbs4--a gain in weight of #b lbs4 #ba4--;,a ,,deal ,,in ,,apples4 ,i was first offered sixteen apples for my shilling1 which would be at the rate of ninepence a dozen4 ,the two extra apples gave me eighteen for a shilling1 which is at the rate of eightpence a dozen1 or one penny a dozen less than the first price asked4 #bb4--;,a ,,deal ,,in ,,eggs4 ,the man must have bought ten eggs at fivepence1 ten eggs at one penny1 and eighty eggs at a halfpenny4 ,he would then have one hundred eggs at a cost of eight shillings and fourpence1 and the same number of eggs of two of the qualities4 #bc4--,,the ,,christmas-,,boxes4 ,the distribution took place some years ago1 when the fourpenny-piece was in circulation4 ,nineteen persons must each have received nineteen pence4 ,there are five different ways in which this sum may have been paid in silver coins4 ,we need only use two of these ways4 ,thus if fourteen men each received four four-penny-pieces and one threepenny-piece1 and five men each received five threepenny-pieces and one fourpenny-piece1 each man would receive nineteen pence1 and there would be exactly one hundred coins of a total value of #a1 #ajs4 #a;d4 #bd4--;,a ,,shopping ,,perplexity4 ,the first purchase amounted to #as4 #e;d41 the second to #as4 #aa;d41 and together they make #cs4 #e;d4 ,not one of these three amounts can be paid in fewer than six current coins of the realm4 #be4--;,,chinese ,,money4 ,as a ching-chang is worth twopence and four-fifteenths of a ching-chang1 the remaining eleven-fifteenths of a ching-chang must be worth twopence4 ,therefore eleven ching-changs are worth exactly thirty pence1 or half a crown4 ,now1 the exchange must be made with seven round-holed coins and one square-holed coin4 ,thus it will be seen that #g round-holed coins are worth seven-elevenths of #ae ching-changs1 and #a square-holed coin is worth one-eleventh of #af ching-changs--that is1 #gg rounds equal #aje ching-changs and #aa squares equal #af ching-changs4 ,therefore #gg rounds added to #aa squares equal #aba ching-changs2 or #g rounds and #a square equal #aa ching-changs1 or its equivalent1 half a crown4 ,this is more simple in practice than it looks here4 #bf4--,,the ,,junior ,,clerks' ,,puzzle4 ,although ,snoggs's reason for wishing to take his rise at #b1 #ajs4 half-yearly did not concern our puzzle1 the fact that he was duping his employer into paying him more than was intended did concern it4 ,many readers will be surprised to find that1 although ,moggs only received #cej in five years1 the artful ,snoggs actually obtained #cfb1 #ajs4 in the same time4 ,the rest is simplicity itself4 ,it is evident that if ,moggs saved #hg1 #ajs4 and ,snoggs #aha1 #es41 the latter would be saving twice as great a proportion of his salary as the former 7namely1 one-half as against one-quarter71 and the two sums added together make #bfh1 #aes4 #bg4--;,,giving ,,change4 ,the way to help the ,american tradesman out of his dilemma is this4 ,describing the coins by the number of cents that they represent1 the tradesman puts on the counter #ej and #be2 the buyer puts down #ajj1 #c1 and #b2 the stranger adds his #aj1 #aj1 #e1 #b1 and #a4 ,now1 considering that the cost of the purchase amounted to #cd cents1 it is clear that out of this pooled money the tradesman has to receive #aji1 the buyer #ga1 and the stranger his #bh cents4 ,therefore it is obvious at a glance that the #ajj-piece must go to the tradesman1 and it then follows that the #ej-piece must go to the buyer1 and then the #be-piece can only go to the stranger4 ,another glance will now make it clear that the two #aj-;cent pieces must go to the buyer1 because the tradesman now only wants #i and the stranger #c4 ,then it becomes obvious that the buyer must take the #a cent1 that the stranger must take the #c cents1 and the tradesman the #e1 #b1 and #b4 ,to sum up1 the tradesman takes #ajj1 #e1 #b1 and #b2 the buyer1 #ej1 #aj1 #aj1 and #a2 the stranger1 #be and #c4 ,it will be seen that not one of the three persons retains any one of his own coins4 #bh4--;,,defective ,,observation4 ,of course the date on a penny is on the same side as ,britannia--the tail side4 ,six pennies may be laid around another penny1 all flat on the table1 so that every one of them touches the central one4 ,the number of threepenny-pieces that may be laid on the surface of a half-crown1 so that no piece lies on another or overlaps the edge of the half-crown1 is one4 ,a second threepenny-piece will overlap the edge of the larger coin4 ,few people guess fewer than three1 and many persons give an absurdly high number4 #bi4--,,the ,,broken ,,coins4 ,if the three broken coins when perfect were worth #bec pence1 and are now in their broken condition worth #bdj pence1 it should be obvious that #ac_/bec of the original value has been lost4 ,and as the same fraction of each coin has been broken away1 each coin has lost #ac_/bec of its original bulk4 #cj4--,,two ,,questions ,,in ,,probabilities4 ,in tossing with the five pennies all at the same time1 it is obvious that there are #cb different ways in which the coins may fall1 because the first coin may fall in either of two ways1 then the second coin may also fall in either of two ways1 and so on4 ,therefore five #b's multiplied together make #cb4 ,now1 how are these #cb ways made up8 ,here they are3-- 7a7 #e heads #a way 7b7 #e tails #a way 7c7 #d heads and #a tail #e ways 7d7 #d tails and #a head #e ways 7e7 #c heads and #b tails #aj ways 7f7 #c tails and #b heads #aj ways ,now1 it will be seen that the only favourable cases are a1 b1 c1 and d--#ab cases4 ,the remaining #bj cases are unfavourable1 because they do not give at least four heads or four tails4 ,therefore the chances are only #ab to #bj in your favour1 or 7which is the same thing7 #c to #e4 ,put another way1 you have only #c chances out of #h4 ,the amount that should be paid for a draw from the bag that contains three sovereigns and one shilling is #aes4 #c;d4 ,many persons will say that1 as one's chances of drawing a sovereign were #c out of #d1 one should pay three-fourths of a pound1 or #aes41 overlooking the fact that one must draw at least a shilling--there being no blanks4 #ca4--;,,domestic ,,economy4 ,without the hint that ,i gave1 my readers would probably have been unanimous in deciding that ,mr4 ,perkins's income must have been #a1gaj4 ,but this is quite wrong4 ,mrs4 ,perkins says1 ,we have spent a third of his yearly income in rent1 etc41 etc4--that is1 in two years they have spent an amount in rent1 etc41 equal to one-third of his yearly income4 ,note that she does not say that they have spent each year this sum1 whatever it is1 but that during the two years that amount has been spent4 ,the only possible answer1 according to the exact reading of her words1 is1 therefore1 that his income was #ahj per annum4 ,thus the amount spent in two years1 during which his income has amounted to #cfj1 will be #fj in rent1 etc41 #ij in domestic expenses1 #bj in other ways1 leaving the balance of #aij in the bank as stated4 #cb4--,,the ,,excursion ,,ticket ,,puzzle4 ,nineteen shillings and ninepence may be paid in #deh1ijh1fbb different ways4 ,i do not propose to give my method of solution4 ,any such explanation would occupy an amount of space out of proportion to its interest or value4 ,if ,i could give within reasonable limits a general solution for all money payments1 ,i would strain a point to find room2 but such a solution would be extremely complex and cumbersome1 and ,i do not consider it worth the labour of working out4 ,just to give an idea of what such a solution would involve1 ,i will merely say that ,i find that1 dealing only with those sums of money that are multiples of threepence1 if we only use bronze coins any sum can be paid in 7n #a7 ways where n always represents the number of pence4 ,if threepenny-pieces are admitted1 there are #bn #aen #ccn --------------------- #a ways4 #ah ,if sixpences are also used there are n#d #bbn #aein #dadn #baf --------------------------------- #baf ways1 when the sum is a multiple of sixpence1 and the constant1 #baf1 changes to #cbd when the money is not such a multiple4 ,and so the formulas increase in complexity in an accelerating ratio as we go on to the other coins4 ,i will1 however1 add an interesting little table of the possible ways of changing our current coins which ,i believe has never been given in a book before4 ,change may be given for a ,farthing in #j way4 ,halfpenny in #a way4 ,penny in #c ways4 ,threepenny-piece in #af ways4 ,sixpence in #ff ways4 ,shilling in #djb ways4 ,florin in #c1hah ways4 ,half-crown in #h1gji ways4 ,double florin in #fj1bci ways4 ,crown in #aff1fea ways4 ,half-sovereign in #f1bfa1fbb ways4 ,sovereign in #ejj1bia1hcc ways4 ,it is a little surprising to find that a sovereign may be changed in over five hundred million different ways4 ,but ,i have no doubt as to the correctness of my figures4 #cc4--;,a ,,puzzle ,,in ,,reversals4 7i7 #ac4 7#b7 #bc1 #ais4 #aa;d4 ,the words the number of pounds exceeds that of the pence exclude such sums of money as #b1 #afs4 #b;d4 and all sums under #a4 #cd4--,,the ,,grocer ,,and ,,draper4 ,the grocer was delayed half a minute and the draper eight minutes and a half 7seventeen times as long as the grocer71 making together nine minutes4 ,now1 the grocer took twenty-four minutes to weigh out the sugar1 and1 with the half-minute delay1 spent #bd min4 #cj sec4 over the task2 but the draper had only to make forty-seven cuts to divide the roll of cloth1 containing forty-eight yards1 into yard pieces6 ,this took him #ae min4 #dj sec41 and when we add the eight minutes and a half delay we get #bd min4 #aj sec41 from which it is clear that the draper won the race by twenty seconds4 ,the majority of solvers make forty-eight cuts to divide the roll into forty-eight pieces6 #ce4--;,,judkins',s ,,cattle4 ,as there were five droves with an equal number of animals in each drove1 the number must be divisible by #e2 and as every one of the eight dealers bought the same number of animals1 the number must be divisible by #h4 ,therefore the number must be a multiple of #dj4 ,the highest possible multiple of #dj that will work will be found to be #abj1 and this number could be made up in one of two ways--#a ox1 #bc pigs1 and #if sheep1 or #c oxen1 #h pigs1 and #aji sheep4 ,but the first is excluded by the statement that the animals consisted of oxen1 pigs1 and sheep1 because a single ox is not oxen4 ,therefore the second grouping is the correct answer4 #cf4--;,,buying ,,apples4 ,as there were the same number of boys as girls1 it is clear that the number of children must be even1 and1 apart from a careful and exact reading of the question1 there would be three different answers4 ,there might be two1 six1 or fourteen children4 ,in the first of these cases there are ten different ways in which the apples could be bought4 ,but we were told there was an equal number of boys and girls1 and one boy and one girl are not boys and girls1 so this case has to be excluded4 ,in the case of fourteen children1 the only possible distribution is that each child receives one halfpenny apple4 ,but we were told that each child was to receive an equal distribution of apples1 and one apple is not apples1 so this case has also to be excluded4 ,we are therefore driven back on our third case1 which exactly fits in with all the conditions4 ,three boys and three girls each receive #a halfpenny apple and #b third-penny apples4 ,the value of these #c apples is one penny and one-sixth1 which multiplied by six makes sevenpence4 ,consequently1 the correct answer is that there were six children--three girls and three boys4 #cg4--;,,buying ,,chestnuts4 ,in solving this little puzzle we are concerned with the exact interpretation of the words used by the buyer and seller4 ,i will give the question again1 this time adding a few words to make the matter more clear4 ,the added words are printed in italics4 ,a man went into a shop to buy chestnuts4 ,he said he wanted a pennyworth1 and was given five chestnuts4 ',it is not enough2 ,i ought to have a sixth of a chestnut more1' he remarked4 ',but if ,i give you one chestnut more1' the shopman replied1 'you will have five-sixths too many4' ,now1 strange to say1 they were both right4 ,how many chestnuts should the buyer receive for half a crown8 ,the answer is that the price was #aee chestnuts for half a crown4 ,divide this number by #cj1 and we find that the buyer was entitled to #ea_/f chestnuts in exchange for his penny4 ,he was1 therefore1 right when he said1 after receiving five only1 that he still wanted a sixth4 ,and the salesman was also correct in saying that if he gave one chestnut more 7that is1 six chestnuts in all7 he would be giving five-sixths of a chestnut in excess4 #ch4--,,the ,,bicycle ,,thief4 ,people give all sorts of absurd answers to this question1 and yet it is perfectly simple if one just considers that the salesman cannot possibly have lost more than the cyclist actually stole4 ,the latter rode away with a bicycle which cost the salesman eleven pounds1 and the ten pounds change2 he thus made off with twenty-one pounds1 in exchange for a worthless bit of paper4 ,this is the exact amount of the salesman's loss1 and the other operations of changing the cheque and borrowing from a friend do not affect the question in the slightest4 ,the loss of prospective profit on the sale of the bicycle is1 of course1 not direct loss of money out of pocket4 #ci4--,,the ,,costermonger',s ,,puzzle4 ,bill must have paid #hs4 per hundred for his oranges--that is1 #abe for #ajs4 ,at #hs4 #d;d4 per hundred1 he would only have received #abj oranges for #ajs4 ,this exactly agrees with ,bill's statement4 #dj4--,,mamma',s ,,age4 ,the age of ,mamma must have been #bi years #b months2 that of ,papa1 #ce years2 and that of the child1 ,tommy1 #e years #aj months4 ,added together1 these make seventy years4 ,the father is six times the age of the son1 and1 after #bc years #d months have elapsed1 their united ages will amount to #adj years1 and ,tommy will be just half the age of his father4 #da4--,,their ,,ages4 ,the gentleman's age must have been #ed years and that of his wife #de years4 #db4--,,the ,,family ,,ages4 ,the ages were as follows3 ,billie1 #c years2 ,gertrude1 #a year2 ,henrietta1 #e years2 ,charlie1 #aj2 years2 and ,janet1 #ba years4 #dc4--,,mrs4 ,,timpkins',s ,,age4 ,the age of the younger at marriage is always the same as the number of years that expire before the elder becomes twice her age1 if he was three times as old at marriage4 ,in our case it was eighteen years afterwards2 therefore ,mrs4 ,timpkins was eighteen years of age on the wedding-day1 and her husband fifty-four4 #dd4--;,a ,,census ,,puzzle4 ,miss ,ada ,jorkins must have been twenty-four and her little brother ,johnnie three years of age1 with thirteen brothers and sisters between4 ,there was a trap for the solver in the words seven times older than little ,johnnie4 ,of course1 seven times older is equal to eight times as old4 ,it is surprising how many people hastily assume that it is the same as seven times as old4 ,some of the best writers have committed this blunder4 ,probably many of my readers thought that the ages #bd and #c were correct4 #de4--,,mother ,,and ,,daughter4 ,in four and a half years1 when the daughter will be sixteen years and a half and the mother forty-nine and a half years of age4 #df4--,,mary ,,and ,,marmaduke4 ,marmaduke's age must have been twenty-nine years and two-fifths1 and ,mary's nineteen years and three-fifths4 ,when ,marmaduke was aged nineteen and three-fifths1 ,mary was only nine and four-fifths2 so ,marmaduke was at that time twice her age4 #dg4--,,rover',s ,,age4 ,rover's present age is ten years and ,mildred's thirty years4 ,five years ago their respective ages were five and twenty-five4 ,remember that we said four times older than the dog1 which is the same as five times as old4 7,see answer to ,no4 #dd47 #dh4--;,,concerning ,,tommy',s ,,age4 ,tommy ,smart's age must have been nine years and three-fifths4 ,ann's age was sixteen and four-fifths1 the mother's thirty-eight and two-fifths1 and the father's fifty and two-fifths4 #di4--,,next-,,door ,,neighbours4 ,mr4 ,jupp #ci1 ,mrs4 ,jupp #cd1 ,julia #ad1 and ,joe #ac2 ,mr4 ,simkin #db2 ,mrs4 ,simkin #dj2 ,sophy #aj2 and ,sammy #h4 #ej4--,,the ,,bag ,,of ,,nuts4 ,it will be found that when ,herbert takes twelve1 ,robert and ,christopher will take nine and fourteen respectively1 and that they will have together taken thirty-five nuts4 ,as #ce is contained in #ggj twenty-two times1 we have merely to multiply #ab1 #i1 and #ad by #bb to discover that ,herbert's share was #bfd1 ,robert's #aih1 and ,christopher's #cjh4 ,then1 as the total of their ages is #ag years or half the sum of #ab1 #i1 and #ad1 their respective ages must be #f1 #d1 and #g years4 #ea4--;,,how ,,old ,,was ,,mary8 ,the age of ,mary to that of ,ann must be as #e to #c4 ,and as the sum of their ages was #dd1 ,mary was #bg and ,ann #af4 ,one is exactly #aa years older than the other4 ,i will now insert in brackets in the original statement the various ages specified3 ,mary is 7#bg7 twice as old as ,ann was 7#ac7 when ,mary was half as old 7#bd7 as ,ann will be 7#di7 when ,ann is three times as old 7#di7 as ,mary was 7#af7 when ,mary was 7#af7 three times as old as ,ann 7#e74 ,now1 check this backwards4 ,when ,mary was three times as old as ,ann1 ,mary was #af and ,ann #e 7#aa years younger74 ,then we get #di for the age ,ann will be when she is three times as old as ,mary was then4 ,when ,mary was half this she was #bd4 ,and at that time ,ann must have been #ac 7#aa years younger74 ,therefore ,mary is now twice as old--#bg1 and ,ann #aa years younger--#af4 #eb4--,,queer ,,relationships4 ,if a man marries a woman1 who dies1 and he then marries his deceased wife's sister and himself dies1 it may be correctly said that he had 7previously7 married the sister of his widow4 ,the youth was not the nephew of ,jane ,brown1 because he happened to be her son4 ,her surname was the same as that of her brother1 because she had married a man of the same name as herself4 #ec4--;,,heard ,,on ,,the ,,tube ,,railway4 ,the gentleman was the second lady's uncle4 #ed4--;,a ,,family ,,party4 ,the party consisted of two little girls and a boy1 their father and mother1 and their father's father and mother4 #ee4--;,a ,,mixed ,,pedigree4 ,illustration3 ,thos4 ,bloggs m 4 4 4 4 4 ------------------------------------ ,w4 ,snoggs m ,kate ,bloggs4 4 4 m ,henry ,bloggs4 ,joseph ,bloggs m --------------------- ,jane ,john ,alf4 ,mary ,bloggs m ,snoggs ,snoggs m ,bloggs ,the letter m stands for married4 ,it will be seen that ,john ,snoggs can say to ,joseph ,bloggs1 ,you are my father's brother-in-law1 because my father married your sister ,kate2 you are my brother's father-in-law1 because my brother ,alfred married your daughter ,mary2 and you are my father-in-law's brother1 because my wife ,jane was your brother ,henry's daughter4 #ef4--,,wilson',s ,,poser4 ,if there are two men1 each of whom marries the mother of the other1 and there is a son of each marriage1 then each of such sons will be at the same time uncle and nephew of the other4 ,there are other ways in which the relationship may be brought about1 but this is the simplest4 #eg4--,,what ,,was ,,the ,,time8 ,the time must have been #i4cf p4m4 ,a quarter of the time since noon is #b hr4 #bd min41 and a half of the time till noon next day is #g hr4 #ab min4 ,these added together make #i hr4 #cf min4 #eh4--;,a ,,time ,,puzzle4 ,twenty-six minutes4 #ei4--;,a ,,puzzling ,,watch4 ,if the #fe minutes be counted on the face of the same watch1 then the problem would be impossible3 for the hands must coincide every #fee_/aa minutes as shown by its face1 and it matters not whether it runs fast or slow2 but if it is measured by true time1 it gains #e_/aa of a minute in #fe minutes1 or #fj_/adc of a minute per hour4 #fj4--,,the ,,wapshaw',s ,,wharf ,,mystery4 ,there are eleven different times in twelve hours when the hour and minute hands of a clock are exactly one above the other4 ,if we divide #ab hours by #aa we get #a hr4 #e min4 #bgc_/aa sec41 and this is the time after twelve o'clock when they are first together1 and also the time that elapses between one occasion of the hands being together and the next4 ,they are together for the second time at #b hr4 #aj min4 #edf_/aa sec4 7twice the above time72 next at #c hr4 #af min4 #bai_/aa sec42 next at #d hr4 #ba min4 #dia_/aa sec4 ,this last is the only occasion on which the two hands are together with the second hand just past the forty-ninth second4 ,this1 then1 is the time at which the watch must have stopped4 ,guy ,boothby1 in the opening sentence of his ,across the ,world for a ,wife1 says1 ,it was a cold1 dreary winter's afternoon1 and by the time the hands of the clock on my mantelpiece joined forces and stood at twenty minutes past four1 my chambers were well-nigh as dark as midnight4 ,it is evident that the author here made a slip1 for1 as we have seen above1 he is #a min4 #dia_/aa sec4 out in his reckoning4 #fa4--;,,changing ,,places4 ,there are thirty-six pairs of times when the hands exactly change places between three p4m4 and midnight4 ,the number of pairs of times from any hour 7n7 to midnight is the sum of #ab - 7n #a7 natural numbers4 ,in the case of the puzzle n #c2 therefore #ab - 7#c #a7 #h and #a #b #c #d #e #f #g #h #cf1 the required answer4 ,the first pair of times is #c hr4 #baeg_/adc min4 and #d hr4 #afaab_/adc min41 and the last pair is #aj hr4 #eihc_/adc min4 and #aa hr4 #edach_/adc min4 ,i will not give all the remainder of the thirty-six pairs of times1 but supply a formula by which any of the sixty-six pairs that occur from midday to midnight may be at once found3-- #gbj;b #fj;a #gbj;a #fj;b min4 a hr ---------- min4 and b hr4 --------------- #adc #adc ,for the letter a may be substituted any hour from #j1 #a1 #b1 #c up to #aj 7where nought stands for #ab o'clock midday72 and b may represent any hour1 later than a1 up to #aa4 ,by the aid of this formula there is no difficulty in discovering the answer to the second question3 a #h and b #aa will give the pair #h hr4 #ehajf_/adc min4 and #aa hr4 #ddabh_/adc min41 the latter being the time when the minute hand is nearest of all to the point ,,ix--in fact1 it is only #ae_/adc of a minute distant4 ,readers may find it instructive to make a table of all the sixty-six pairs of times when the hands of a clock change places4 ,an easy way is as follows3 ,make a column for the first times and a second column for the second times of the pairs4 ,by making a #j and b #a in the above expressions we find the first case1 and enter hr4 #ee_/adc min4 at the head of the first column1 and #a hr4 #jfj_/adc min4 at the head of the second column4 ,now1 by successively adding #ee_/adc min4 in the first1 and #a hr4 #jfj_/adc min4 in the second column1 we get all the eleven pairs in which the first time is a certain number of minutes after nought1 or mid-day4 ,then there is a jump in the times1 but you can find the next pair by making a #a and b #b1 and then by successively adding these two times as before you will get all the ten pairs after #a o'clock4 ,then there is another jump1 and you will be able to get by addition all the nine pairs after #b o'clock4 ,and so on to the end4 ,i will leave readers to investigate for themselves the nature and cause of the jumps4 ,in this way we get under the successive hours1 #aa #aj #i #h #g #f #e #d #c #b #a #ff pairs of times1 which result agrees with the formula in the first paragraph of this article4 ,some time ago the principal of a ,civil ,service ,training ,college1 who conducts a ,civil ,service ,column in one of the periodicals1 had the query addressed to him1 ,how soon after ,,xii o'clock will a clock with both hands of the same length be ambiguous8 ,his first answer was1 ,some time past one o'clock1 but he varied the answer from issue to issue4 ,at length some of his readers convinced him that the answer is1 ,at #ee_/adc min4 past ,,xii2 and this he finally gave as correct1 together with the reason for it that at that time the time indicated is the same whichever hand you may assume as hour hand6 #fb4--,,the ,,club ,,clock4 ,the positions of the hands shown in the illustration could only indicate that the clock stopped at #dd min4 #eaaadc_/adbg sec4 after eleven o'clock4 ,the second hand would next be exactly midway between the other two hands at #de min4 #ebdif_/adbg sec4 after eleven o'clock4 ,if we had been dealing with the points on the circle to which the three hands are directed1 the answer would be #de min4 #bbajf_/adbg sec4 after eleven2 but the question applied to the hands1 and the second hand would not be between the others at that time1 but outside them4 #fc4--,,the ,,stop-,,watch4 ,the time indicated on the watch was #ee_/aa min4 past #i1 when the second hand would be at #bgc_/aa sec4 ,the next time the hands would be similar distances apart would be #edf_/aa min4 past #b1 when the second hand would be at #cbh_/aa sec4 ,but you need only hold the watch 7or our previous illustration of it7 in front of a mirror1 when you will see the second time reflected in it6 ,of course1 when reflected1 you will read ,,xi as ,i1 ,x as ,,ii1 and so on4 #fd4--,,the ,,three ,,clocks4 ,as a mere arithmetical problem this question presents no difficulty4 ,in order that the hands shall all point to twelve o'clock at the same time1 it is necessary that ,b shall gain at least twelve hours and that ,c shall lose twelve hours4 ,as ,b gains a minute in a day of twenty-four hours1 and ,c loses a minute in precisely the same time1 it is evident that one will have gained #gbj minutes 7just twelve hours7 in #gbj days1 and the other will have lost #gbj minutes in #gbj days4 ,clock ,a keeping perfect time1 all three clocks must indicate twelve o'clock simultaneously at noon on the #gbjth day from ,april #a1 #ahih4 ,what day of the month will that be8 ,i published this little puzzle in #ahih to see how many people were aware of the fact that #aijj would not be a leap year4 ,it was surprising how many were then ignorant on the point4 ,every year that can be divided by four without a remainder is bissextile or leap year1 with the exception that one leap year is cut off in the century4 #ahjj was not a leap year1 nor was #aijj4 ,on the other hand1 however1 to make the calendar more nearly agree with the sun's course1 every fourth hundred year is still considered bissextile4 ,consequently1 #bjjj1 #bdjj1 #bhjj1 #cbjj1 etc41 will all be leap years4 ,may my readers live to see them4 ,we therefore find that #gbj days from noon of ,april #a1 #ahih1 brings us to noon of ,march #bb1 #aijj4 #fe4--,,the ,,railway ,,station ,,clock4 ,the time must have been #dcg_/aa min4 past two o'clock4 #ff4--,,the ,,village ,,simpleton4 ,the day of the week on which the conversation took place was ,sunday4 ,for when the day after to-morrow 7,tuesday7 is yesterday1 to-day will be ,wednesday2 and when the day before yesterday 7,friday7 was to-morrow1 to-day was ,thursday4 ,there are two days between ,thursday and ,sunday1 and between ,sunday and ,wednesday4 #fg4--;,,average ,,speed4 ,the average speed is twelve miles an hour1 not twelve and a half1 as most people will hastily declare4 ,take any distance you like1 say sixty miles4 ,this would have taken six hours going and four hours returning4 ,the double journey of #abj miles would thus take ten hours1 and the average speed is clearly twelve miles an hour4 #fh4--,,the ,,two ,,trains4 ,one train was running just twice as fast as the other4 #fi4--,,the ,,three ,,villages4 ,calling the three villages by their initial letters1 it is clear that the three roads form a triangle1 ,a1 ,b1 ,c1 with a perpendicular1 measuring twelve miles1 dropped from ,c to the base ,a1 ,b4 ,this divides our triangle into two right-angled triangles with a twelve-mile side in common4 ,it is then found that the distance from ,a to ,c is #ae miles1 from ,c to ,b #bj miles1 and from ,a to ,b #be 7that is #i and #af7 miles4 ,these figures are easily proved1 for the square of #ab added to the square of #i equals the square of #ae1 and the square of #ab added to the square of #af equals the square of #bj4 #gj4--;,,drawing ,,her ,,pension4 ,the distance must be #f miles4 #ga4--,,sir ,,edwyn ,,de ,,tudor4 ,the distance must have been sixty miles4 ,if ,sir ,edwyn left at noon and rode #ae miles an hour1 he would arrive at four o'clock--an hour too soon4 ,if he rode #aj miles an hour1 he would arrive at six o'clock--an hour too late4 ,but if he went at #ab miles an hour1 he would reach the castle of the wicked baron exactly at five o'clock--the time appointed4 #gb4--,,the ,,hydroplane ,,question4 ,the machine must have gone at the rate of seven-twenty-fourths of a mile per minute and the wind travelled five-twenty-fourths of a mile per minute4 ,thus1 going1 the wind would help1 and the machine would do twelve-twenty-fourths1 or half a mile a minute1 and returning only two-twenty-fourths1 or one-twelfth of a mile per minute1 the wind being against it4 ,the machine without any wind could therefore do the ten miles in thirty-four and two-sevenths minutes1 since it could do seven miles in twenty-four minutes4 #gc4--;,,donkey ,,riding4 ,the complete mile was run in nine minutes4 ,from the facts stated we cannot determine the time taken over the first and second quarter-miles separately1 but together they1 of course1 took four and a half minutes4 ,the last two quarters were run in two and a quarter minutes each4 #gd4--,,the ,,basket ,,of ,,potatoes4 ,multiply together the number of potatoes1 the number less one1 and twice the number less one1 then divide by #c4 ,thus #ej1 #di1 and #ii multiplied together make #bdb1eej1 which1 divided by #c1 gives us #hj1hej yards as the correct answer4 ,the boy would thus have to travel #de miles and fifteen-sixteenths--a nice little recreation after a day's work4 #ge4--,,the ,,passenger',s ,,fare4 ,mr4 ,tompkins should have paid fifteen shillings as his correct share of the motor-car fare4 ,he only shared half the distance travelled for #c1 and therefore should pay half of thirty shillings1 or fifteen shillings4 #gf4--,,the ,,barrel ,,of ,,beer4 ,here the digital roots of the six numbers are #f1 #d1 #a1 #b1 #g1 #i1 which together sum to #bi1 whose digital root is #b4 ,as the contents of the barrels sold must be a number divisible by #c1 if one buyer purchased twice as much as the other1 we must find a barrel with root #b1 #e1 or #h to set on one side4 ,there is only one barrel1 that containing #bj gallons1 that fulfils these conditions4 ,so the man must have kept these #bj gallons of beer for his own use and sold one man #cc gallons 7the #ah-;gallon and #ae-;gallon barrels7 and sold the other man #ff gallons 7the #af1 #ai1 and #ca gallon barrels74 #gg4--;,,digits ,,and ,,squares4 ,the top row must be one of the four following numbers3 #aib1 #bai1 #bgc1 #cbg4 ,the first was the example given4 #gh4--,,odd ,,and ,,even ,,digits4 ,as we have to exclude complex and improper fractions and recurring decimals1 the simplest solution is this3 #gi #ea_/c and #hd #b_/f1 both equal #hda_/c4 ,without any use of fractions it is obviously impossible4 #gi4--,,the ,,lockers ,,puzzle4 ,the smallest possible total is #cef #ajg #bdi1 and the largest sum possible is #iha #bce #gdf1 or #fegcbd4 ,the middle sum may be either #gbj #acd #ehf1 or #gjb #acd #efh1 or #djg #ach #bfi4 ,the total in this case must be made up of three of the figures #j1 #b1 #d1 #g1 but no sum other than the three given can possibly be obtained4 ,we have therefore no choice in the case of the first locker1 an alternative in the case of the third1 and any one of three arrangements in the case of the middle locker4 ,here is one solution3-- #ajg #acd #bce #bdi #ehf #gdf --- --- --- #cef #gbj #iha ,of course1 in each case figures in the first two lines may be exchanged vertically without altering the total1 and as a result there are just #c1jgb different ways in which the figures might be actually placed on the locker doors4 ,i must content myself with showing one little principle involved in this puzzle4 ,the sum of the digits in the total is always governed by the digit omitted4 #i_/i - #g_/aj - #e_/aa -#c_/ab - #a_/ac - #h_/ad - #f_/ae - #d_/af - #b_/ag - #j_/ah4 ,whichever digit shown here in the upper line we omit1 the sum of the digits in the total will be found beneath it4 ,thus in the case of locker ,a we omitted #h1 and the figures in the total sum up to #ad4 ,if1 therefore1 we wanted to get #cef1 we may know at once to a certainty that it can only be obtained 7if at all7 by dropping the #h4 #hj4--,,the ,,three ,,groups4 ,there are nine solutions to this puzzle1 as follows1 and no more3-- #ab #dhc #e1gif #bg #aih #e1cdf #db #ach #e1gif #ci #ahf #g1bed #ah #big #e1cdf #dh #aei #g1fcb #bh #aeg #d1cif #d #a1gch #f1ieb #d #a1ifc #g1heb ,the seventh answer is the one that is most likely to be overlooked by solvers of the puzzle4 #ha4--,,the ,,nine ,,counters4 ,in this case a certain amount of mere trial is unavoidable4 ,but there are two kinds of trials--those that are purely haphazard1 and those that are methodical4 ,the true puzzle lover is never satisfied with mere haphazard trials4 ,the reader will find that by just reversing the figures in #bc and #df 7making the multipliers #cb and #fd7 both products will be #e1jef4 ,this is an improvement1 but it is not the correct answer4 ,we can get as large a product as #e1efh if we multiply #agd by #cb and #if by #eh1 but this solution is not to be found without the exercise of some judgment and patience4 #hb4--,,the ,,ten ,,counters4 ,as ,i pointed out1 it is quite easy so to arrange the counters that they shall form a pair of simple multiplication sums1 each of which will give the same product--in fact1 this can be done by anybody in five minutes with a little patience4 ,but it is quite another matter to find that pair which gives the largest product and that which gives the smallest product4 ,now1 in order to get the smallest product1 it is necessary to select as multipliers the two smallest possible numbers4 ,if1 therefore1 we place #a and #b as multipliers1 all we have to do is to arrange the remaining eight counters in such a way that they shall form two numbers1 one of which is just double the other2 and in doing this we must1 of course1 try to make the smaller number as low as possible4 ,of course the lowest number we could get would be #c1jde2 but this will not work1 neither will #c1dje1 #c1de,o1 etc41 and it may be ascertained that #c1dhe is the lowest possible4 ,one of the required answers is #c1dhe #b #f1igj1 and #f1igj #a #f1igj4 ,the other part of the puzzle 7finding the pair with the highest product7 is1 however1 the real knotty point1 for it is not at all easy to discover whether we should let the multiplier consist of one or of two figures1 though it is clear that we must keep1 so far as we can1 the largest figures to the left in both multiplier and multiplicand4 ,it will be seen that by the following arrangement so high a number as #eh1efj may be obtained4 ,thus1 #iae #fd #eh1efj1 and #gcb #hj #eh1efj4 #hc4--;,,digital ,,multiplication4 ,the solution that gives the smallest possible sum of digits in the common product is #bc #agd #eh #fi #d1jjb1 and the solution that gives the largest possible sum of digits1 #ifed #ahcbge1hhf4 ,in the first case the digits sum to #f and in the second case to #bg4 ,there is no way of obtaining the solution but by actual trial4 #hd4--,,the ,,pierrot',s ,,puzzle4 ,there are just six different solutions to this puzzle1 as follows3-- #h multiplied by #dgc equals #cghd #i #cea #caei #ae #ic #acie #ba #hg #abhg #bg #ha #bahg #ce #da #adce ,it will be seen that in every case the two multipliers contain exactly the same figures as the product4 #he4--,,the ,,cab ,,numbers4 ,the highest product is1 ,i think1 obtained by multiplying #h1gde1bca by #if--namely1 #hci1edb1agf4 ,dealing here with the problem generally1 ,i have shown in the last puzzle that with three digits there are only two possible solutions1 and with four digits only six different solutions4 ,these cases have all been given4 ,with five digits there are just twenty-two solutions1 as follows3-- #c #dabh #abchd #c #dbha #abhdc #c #gabe #bacge #c #gbea #bagec #beda #f #aebdf #fea #bd #aefbd #fgh #db #bhdgf #bdf #ea #abedf #eg #hcd #dgech #ge #bca #agcbe #fbd #gh #dhfgb #dce #hg #cghde ------ #i #gdfa #fgadi #gb #icf #fgcib ------ #b #hgad #agdbh #b #hgda #agdhb #fe #bha #ahbfe #fe #ihc #fcihe ------ #digc #h #cighd #feba #h #ebafh #ad #ibf #abifd #hf #bea #baehf ,now1 if we took every possible combination and tested it by multiplication1 we should need to make no fewer than #cj1bdj trials1 or1 if we at once rejected the number #a as a multiplier1 #bh1efj trials--a task that ,i think most people would be inclined to shirk4 ,but let us consider whether there be no shorter way of getting at the results required4 ,i have already explained that if you add together the digits of any number and then1 as often as necessary1 add the digits of the result1 you must ultimately get a number composed of one figure4 ,this last number ,i call the digital root4 ,it is necessary in every solution of our problem that the root of the sum of the digital roots of our multipliers shall be the same as the root of their product4 ,there are only four ways in which this can happen3 when the digital roots of the multipliers are #c and #f1 or #i and #i1 or #b and #b1 or #e and #h4 ,i have divided the twenty-two answers above into these four classes4 ,it is thus evident that the digital root of any product in the first two classes must be #i1 and in the second two classes #d4 ,owing to the fact that no number of five figures can have a digital sum less than #ae or more than #ce1 we find that the figures of our product must sum to either #ah or #bg to produce the root #i1 and to either #bb or #ca to produce the root #d4 ,there are #c ways of selecting five different figures that add up to #ah1 there are #aa ways of selecting five figures that add up to #bg1 there are #i ways of selecting five figures that add up to #bb1 and #e ways of selecting five figures that add up to #ca4 ,there are1 therefore1 #bh different groups1 and no more1 from any one of which a product may be formed4 ,we next write out in a column these #bh sets of five figures1 and proceed to tabulate the possible factors1 or multipliers1 into which they may be split4 ,roughly speaking1 there would now appear to be about #b1jjj possible cases to be tried1 instead of the #cj1bdj mentioned above2 but the process of elimination now begins1 and if the reader has a quick eye and a clear head he can rapidly dispose of the large bulk of these cases1 and there will be comparatively few test multiplications necessary4 ,it would take far too much space to explain my own method in detail1 but ,i will take the first set of figures in my table and show how easily it is done by the aid of little tricks and dodges that should occur to everybody as he goes along4 ,my first product group of five figures is #hd1cba4 ,here1 as we have seen1 the root of each factor must be #c or a multiple of #c4 ,as there is no #f or #i1 the only single multiplier is #c4 ,now1 the remaining four figures can be arranged in #bd different ways1 but there is no need to make #bd multiplications4 ,we see at a glance that1 in order to get a five-figure product1 either the #h or the #d must be the first figure to the left4 ,but unless the #b is preceded on the right by the #h1 it will produce when multiplied either a #f or a #g1 which must not occur4 ,we are1 therefore1 reduced at once to the two cases1 #c #d1abh and #c x #d1bha1 both of which give correct solutions4 ,suppose next that we are trying the two-figure factor1 #ba4 ,here we see that if the number to be multiplied is under #ejj the product will either have only four figures or begin with #aj4 ,therefore we have only to examine the cases #ba #hdc and #ba #hcd4 ,but we know that the first figure will be repeated1 and that the second figure will be twice the first figure added to the second4 ,consequently1 as twice #c added to #d produces a nought in our product1 the first case is at once rejected4 ,it only remains to try the remaining case by multiplication1 when we find it does not give a correct answer4 ,if we are next trying the factor #ab1 we see at the start that neither the #h nor the #c can be in the units place1 because they would produce a #f1 and so on4 ,a sharp eye and an alert judgment will enable us thus to run through our table in a much shorter time than would be expected4 ,the process took me a little more than three hours4 ,i have not attempted to enumerate the solutions in the cases of six1 seven1 eight1 and nine digits1 but ,i have recorded nearly fifty examples with nine digits alone4 #hf4--,,queer ,,multiplication4 ,if we multiply #cbedghia by #f1 we get the product1 #aiebhgcdf4 ,in both cases all the nine digits are used once and once only4 #hg4--,,the ,,number ,,checks ,,puzzle4 ,divide the ten checks into the following three groups3 #g #a #e--d #f--c #b #h #i #j1 and the first multiplied by the second produces the third4 #hh4--;,,digital ,,division4 ,it is convenient to consider the digits as arranged to form fractions of the respective values1 one-half1 one-third1 one-fourth1 one-fifth1 one-sixth1 one-seventh1 one-eighth1 and one-ninth4 ,i will first give the eight answers1 as follows3-- #fgbi_/acdeh #a_/b #ehbc_/agdfi #a_/c #cidb_/aegfh #a_/d #bfig_/acdhe #a_/e #bidc_/agfeh #a_/f #bcid_/afgeh #a_/g #cahg_/bedif #a_/h #fcha_/egdbi #a_/i ,the sum of the numerator digits and the denominator digits will1 of course1 always be #de1 and the digital root is #i4 ,now1 if we separate the nine digits into any two groups1 the sum of the two digital roots will always be #i4 ,in fact1 the two digital roots must be either #i--i1 #h--a1 #g--b1 #f--c1 or #e--d4 ,in the first case the actual sum is #ah1 but then the digital root of this number is itself #i4 ,the solutions in the cases of one-third1 one-fourth1 one-sixth1 one-seventh1 and one-ninth must be of the form #i--i2 that is to say1 the digital roots of both numerator and denominator will be #i4 ,in the cases of one-half and one-fifth1 however1 the digital roots are #f--c1 but of course the higher root may occur either in the numerator or in the denominator2 thus #bfig_/acdhe1 #bgfi_/achde1 #bigc_/adhfe1 #cgbi_/ahfde1 where1 in the first two arrangements1 the roots of the numerator and denominator are respectively #f--c1 and in the last two #c--f4 ,the most curious case of all is1 perhaps1 one-eighth1 for here the digital roots may be of any one of the five forms given above4 ,the denominators of the fractions being regarded as the numerators multiplied by #b1 #c1 #d1 #e1 #f1 #g1 #h1 and #i respectively1 we must pay attention to the carryings over4 ,in order to get five figures in the product there will1 of course1 always be a carry-over after multiplying the last figure to the left1 and in every case higher than #d we must carry over at least three times4 ,consequently in cases from one-fifth to one-ninth we cannot produce different solutions by a mere change of position of pairs of figures1 as1 for example1 we may with #ehcb_/agdif and #ehbc_/agdfi1 where the #b_/f and #c_/i change places4 ,it is true that the same figures may often be differently arranged1 as shown in the two pairs of values for one-fifth that ,i have given in the last paragraph1 but here it will be found there is a general readjustment of figures and not a simple changing of the positions of pairs4 ,there are other little points that would occur to every solver--such as that the figure #e cannot ever appear to the extreme right of the numerator1 as this would result in our getting either a nought or a second #e in the denominator4 ,similarly #a cannot ever appear in the same position1 nor #f in the fraction one-sixth1 nor an even figure in the fraction one-fifth1 and so on4 ,the preliminary consideration of such points as ,i have touched upon will not only prevent our wasting a lot of time in trying to produce impossible forms1 but will lead us more or less directly to the desired solutions4 #hi4--;,,adding ,,the ,,digits4 ,the smallest possible sum of money is #a1 #hs4 #i;d41 the digits of which add to #be4 #ij4--,,the ,,century ,,puzzle4 ,the problem of expressing the number #ajj as a mixed number or fraction1 using all the nine digits once1 and once only1 has1 like all these digital puzzles1 a fascinating side to it4 ,the merest tyro can by patient trial obtain correct results1 and there is a singular pleasure in discovering and recording each new arrangement akin to the delight of the botanist in finding some long-sought plant4 ,it is simply a matter of arranging those nine figures correctly1 and yet with the thousands of possible combinations that confront us the task is not so easy as might at first appear1 if we are to get a considerable number of results4 ,here are eleven answers1 including the one ,i gave as a specimen3-- #badh #ageb #adbh #aegh #if ----1 #if ----1 #if ----1 #id ----1 #ecg #dch #ceg #bfc #gebd #ehbc #egdb #cedf #ia ----1 #ia ----1 #ia ----1 #hb ----1 #hcf #fdg #fch #aig #gebd #efdc #fibeh #ha ----1 #ha ----1 #c -----4 #cif #big #gad ,now1 as all the fractions necessarily represent whole numbers1 it will be convenient to deal with them in the following form3 #if #d1 #id #f1 #ia #i1 #hb #ah1 #ha #ai1 and #c #ig4 ,with any whole number the digital roots of the fraction that brings it up to #ajj will always be of one particular form4 ,thus1 in the case of #if #d1 one can say at once that if any answers are obtainable1 then the roots of both the numerator and the denominator of the fraction will be #f4 ,examine the first three arrangements given above1 and you will find that this is so4 ,in the case of #id #f the roots of the numerator and denominator will be respectively #c--b1 in the case of #ia #i and of #hb #ah they will be #i--h1 in the case of #ha #ai they will be #i--i1 and in the case of #c #ig they will be #c--c4 ,every fraction that can be employed has1 therefore1 its particular digital root form1 and you are only wasting your time in unconsciously attempting to break through this law4 ,every reader will have perceived that certain whole numbers are evidently impossible4 ,thus1 if there is a #e in the whole number1 there will also be a nought or a second #e in the fraction1 which are barred by the conditions4 ,then multiples of #aj1 such as #ij and #hj1 cannot of course occur1 nor can the whole number conclude with a #i1 like #hi and #gi1 because the fraction1 equal to #aa or #ba1 will have #a in the last place1 and will therefore repeat a figure4 ,whole numbers that repeat a figure1 such as #hh and #gg1 are also clearly useless4 ,these cases1 as ,i have said1 are all obvious to every reader4 ,but when ,i declare that such combinations as #ih #b1 #ib #h1 #hf #ad1 #hc #ag1 #gd #bf1 etc41 etc41 are to be at once dismissed as impossible1 the reason is not so evident1 and ,i unfortunately cannot spare space to explain it4 ,but when all those combinations have been struck out that are known to be impossible1 it does not follow that all the remaining possible forms will actually work4 ,the elemental form may be right enough1 but there are other and deeper considerations that creep in to defeat our attempts4 ,for example1 #ih #b is an impossible combination1 because we are able to say at once that there is no possible form for the digital roots of the fraction equal to #b4 ,but in the case of #ig #c there is a possible form for the digital roots of the fraction1 namely1 #f--e1 and it is only on further investigation that we are able to determine that this form cannot in practice be obtained1 owing to curious considerations4 ,the working is greatly simplified by a process of elimination1 based on such considerations as that certain multiplications produce a repetition of figures1 and that the whole number cannot be from #ab to #bc inclusive1 since in every such case sufficiently small denominators are not available for forming the fractional part4 #ia4--,,more ,,mixed ,,fractions4 ,the point of the present puzzle lies in the fact that the numbers #ae and #ah are not capable of solution4 ,there is no way of determining this without trial4 ,here are answers for the ten possible numbers3-- #iedgb_/acfh #ac2 #ifdce_/abhg #ad2 #abcegf_/hid #af2 #facbeh_/idg #bj2 #aeidcb_/ghf #bg2 #bdigef_/hac #cf2 #bgeadh_/cif #dj2 #feahib_/dgc #fi2 #eicfad_/bgh #gb2 #gecfdh_/aib #id4 ,i have only found the one arrangement for each of the numbers #af1 #bj1 and #bg2 but the other numbers are all capable of being solved in more than one way4 ,as for #ae and #ah1 though these may be easily solved as a simple fraction1 yet a mixed fraction assumes the presence of a whole number2 and though my own idea for dodging the conditions is the following1 where the fraction is both complex and mixed1 it will be fairer to keep exactly to the form indicated3-- #cieb ---- #gdf #ae2 #c ---- #a #egdb ---- #fch #ah4 #i ---- #a ,i have proved the possibility of solution for all numbers up to #ajj1 except #a1 #b1 #c1 #d1 #ae1 and #ah4 ,the first three are easily shown to be impossible4 ,i have also noticed that numbers whose digital root is #h--such as #bf1 #ce1 #dd1 #ec1 etc4--seem to lend themselves to the greatest number of answers4 ,for the number #bf alone ,i have recorded no fewer than twenty-five different arrangements1 and ,i have no doubt that there are many more4 #ib4--;,,digital ,,square ,,numbers4 ,so far as ,i know1 there are no published tables of square numbers that go sufficiently high to be available for the purposes of this puzzle4 ,the lowest square number containing all the nine digits once1 and once only1 is #aci1hed1bgf1 the square of #aa1hbf4 ,the highest square number under the same conditions is1 #ibc1ahg1def1 the square of #cj1chd4 #ic4--,,the ,,mystic ,,eleven4 ,most people know that if the sum of the digits in the odd places of any number is the same as the sum of the digits in the even places1 then the number is divisible by #aa without remainder4 ,thus in #hifgdcjab the odd digits1 #bjdfh1 add up #bj1 and the even digits1 #acgi1 also add up #bj4 ,therefore the number may be divided by #aa4 ,but few seem to know that if the difference between the sum of the odd and the even digits is #aa1 or a multiple of #aa1 the rule equally applies4 ,this law enables us to find1 with a very little trial1 that the smallest number containing nine of the ten digits 7calling nought a digit7 that is divisible by #aa is #ajb1cdg1ehf1 and the highest number possible1 #ihg1feb1dac4 #id4--,,the ,,digital ,,century4 ,there is a very large number of different ways in which arithmetical signs may be placed between the nine digits1 arranged in numerical order1 so as to give an expression equal to #ajj4 ,in fact1 unless the reader investigated the matter very closely1 he might not suspect that so many ways are possible4 ,it was for this reason that ,i added the condition that not only must the fewest possible signs be used1 but also the fewest possible strokes4 ,in this way we limit the problem to a single solution1 and arrive at the simplest and therefore 7in this case7 the best result4 ,just as in the case of magic squares there are methods by which we may write down with the greatest ease a large number of solutions1 but not all the solutions1 so there are several ways in which we may quickly arrive at dozens of arrangements of the ,digital ,century1 without finding all the possible arrangements4 ,there is1 in fact1 very little principle in the thing1 and there is no certain way of demonstrating that we have got the best possible solution4 ,all ,i can say is that the arrangement ,i shall give as the best is the best ,i have up to the present succeeded in discovering4 ,i will give the reader a few interesting specimens1 the first being the solution usually published1 and the last the best solution that ,i know4 ,signs4 ,strokes4 #a #b #c #d #e #f #g 7#h #i7 #ajj 7 #i #ah7 - 7#a #b7 - #c - #d - #e 7#f #g7 7#h #i7 #ajj 7#ab #bj7 #a 7#b #c7 7#d #e7 - #f #g 7#h #i7 #ajj 7#aa #ba7 7#a #b - #c - #d77e - #f - #g - #h - #i7 #ajj 7 #i #ab7 #a 7#b #c7 #d #e #fg #h #i #ajj 7#h #af7 7#a #b7 #cd #ef #g - #h #i #ajj 7#g #ac7 #ab #c - #d #e #fg #h #i #ajj 7#f #aa7 #abc - #d - #e - #f - #g #h - #i #ajj 7#f #g7 #abc #d - #e #fg - #h - #i #ajj 7#d #f7 #abc #de - #fg #h - #i #ajj 7#d #f7 #abc - #de - #fg #hi #ajj 7#c #d7 ,it will be noticed that in the above ,i have counted the bracket as one sign and two strokes4 ,the last solution is singularly simple1 and ,i do not think it will ever be beaten4 #ie4--,,the ,,four ,,sevens4 ,the way to write four sevens with simple arithmetical signs so that they represent #ajj is as follows3-- #g #g -- -- #ajj4 4#g 4#g ,of course the fraction1 #g over decimal #g1 equals #g divided by #g_/aj1 which is the same as #gj divided by #g1 or #aj4 ,then #aj multiplied by #aj is #ajj1 and there you are6 ,it will be seen that this solution applies equally to any number whatever that you may substitute for #g4 #if4--,,the ,,dice ,,numbers4 ,the sum of all the numbers that can be formed with any given set of four different figures is always #f1fff multiplied by the sum of the four figures4 ,thus1 #a1 #b1 #c1 #d add up #aj1 and ten times #f1fff is #ff1ffj4 ,now1 there are thirty-five different ways of selecting four figures from the seven on the dice--remembering the #f and #i trick4 ,the figures of all these thirty-five groups add up to #fjj4 ,therefore #f1fff multiplied by #fjj gives us #c1iii1fjj as the correct answer4 ,let us discard the dice and deal with the problem generally1 using the nine digits1 but excluding nought4 ,now1 if you were given simply the sum of the digits--that is1 if the condition were that you could use any four figures so long as they summed to a given amount--then we have to remember that several combinations of four digits will1 in many cases1 make the same sum4 #aj #aa #ab #ac #ad #ae #af #ag #ah #ai #bj #a #a #b #c #e #f #h #i #aa #aa #ab #ba #bb #bc #bd #be #bf #bg #bh #bi #cj #aa #aa #i #h #f #e #c #b #a #a ,here the top row of numbers gives all the possible sums of four different figures1 and the bottom row the number of different ways in which each sum may be made4 ,for example #ac may be made in three ways3 #abcg1 #abdf1 and #acde4 ,it will be found that the numbers in the bottom row add up to #abf1 which is the number of combinations of nine figures taken four at a time4 ,from this table we may at once calculate the answer to such a question as this3 ,what is the sum of all the numbers composed of our different digits 7nought excluded7 that add up to #ad8 ,multiply #ad by the number beneath t in the table1 #e1 and multiply the result by #f1fff1 and you will have the answer4 ,it follows that1 to know the sum of all the numbers composed of four different digits1 if you multiply all the pairs in the two rows and then add the results together1 you will get #b1ebj1 which1 multiplied by #f1fff1 gives the answer #af1gih1cbj4 ,the following general solution for any number of digits will doubtless interest readers4 ,let n represent number of digits1 then #e 7#ajn - #a7 #h6 divided by 7#i - n76 equals the required sum4 ,note that #j6 equals #a4 ,this may be reduced to the following practical rule3 ,multiply together #d #g #f #e 444 to 7n - #a7 factors2 now add 7n #a7 ciphers to the right1 and from this result subtract the same set of figures with a single cipher to the right4 ,thus for n #d 7as in the case last mentioned71 #d #g #f #afh4 ,therefore #af1hjj1jjj less #a1fhj gives us #af1gih1cbj in another way4 #ig4--,,the ,,spot ,,on ,,the ,,table4 ,the ordinary schoolboy would correctly treat this as a quadratic equation4 ,here is the actual arithmetic4 ,double the product of the two distances from the walls4 ,this gives us #add1 which is the square of #ab4 ,the sum of the two distances is #ag4 ,if we add these two numbers1 #ab and #ag1 together1 and also subtract one from the other1 we get the two answers that #bi or #e was the radius1 or half-diameter1 of the table4 ,consequently1 the full diameter was #eh in4 or #aj in4 ,but a table of the latter dimensions would be absurd1 and not at all in accordance with the illustration4 ,therefore the table must have been #eh in4 in diameter4 ,in this case the spot was on the edge nearest to the corner of the room--to which the boy was pointing4 ,if the other answer were admissible1 the spot would be on the edge farthest from the corner of the room4 #ih4--;,,academic ,,courtesies4 ,there must have been ten boys and twenty girls4 ,the number of bows girl to girl was therefore #chj1 of boy to boy #ij1 of girl with boy #djj1 and of boys and girls to teacher #cj1 making together #ijj1 as stated4 ,it will be remembered that it was not said that the teacher himself returned the bows of any child4 #ii4--,,the ,,thirty-,,three ,,pearls4 ,the value of the large central pearl must have been #c1jjj4 ,the pearl at one end 7from which they increased in value by #ajj7 was #a1djj2 the pearl at the other end1 #fjj4 #ajj4--,,the ,,labourer',s ,,puzzle4 ,the man said1 ,i am going twice as deep1 not as deep again4 ,that is to say1 he was still going twice as deep as he had gone already1 so that when finished the hole would be three times its present depth4 ,then the answer is that at present the hole is #c ft4 #f in4 deep and the man #b ft4 #d in4 above ground4 ,when completed the hole will be #aj ft4 #f in4 deep1 and therefore the man will then be #d ft4 #h in4 below the surface1 or twice the distance that he is now above ground4 #aja4--,,the ,,trusses ,,of ,,hay4 ,add together the ten weights and divide by #d1 and we get #bhi lbs4 as the weight of the five trusses together4 ,if we call the five trusses in the order of weight ,a1 ,b1 ,c1 ,d1 and ,e1 the lightest being ,a and the heaviest ,e1 then the lightest1 no lbs41 must be the weight of ,a and ,b2 and the next lightest1 #aab lbs41 must be the weight of ,a and ,c4 ,then the two heaviest1 ,d and ,e1 must weigh #aba lbs41 and ,c and ,e must weigh #abj lbs4 ,we thus know that ,a1 ,b1 ,d1 and ,e weigh together #bca lbs41 which1 deducted from #bhi lbs4 7the weight of the five trusses71 gives us the weight of ,c as #eh lbs4 ,now1 by mere subtraction1 we find the weight of each of the five trusses--#ed lbs41 #ef lbs41 #eh lbs41 #ei lbs41 and #fb lbs4 respectively4 #ajb4--,,mr4 ,,gubbins ,,in ,a ,,fog4 ,the candles must have burnt for three hours and three-quarters4 ,one candle had one-sixteenth of its total length left and the other four-sixteenths4 #ajc4--,,painting ,,the ,,lamp-,,posts4 ,pat must have painted six more posts than ,tim1 no matter how many lamp-posts there were4 ,for example1 suppose twelve on each side2 then ,pat painted fifteen and ,tim nine4 ,if a hundred on each side1 ,pat painted one hundred and three1 and ,tim only ninety-seven #ajd4--;,,catching ,,the ,,thief4 ,the constable took thirty steps4 ,in the same time the thief would take forty-eight1 which1 added to his start of twenty-seven1 carried him seventy-five steps4 ,this distance would be exactly equal to thirty steps of the constable4 #aje4--,,the ,,parish ,,council ,,election1 ,the voter can vote for one candidate in #bc ways1 for two in #bec ways1 for three in #a1gga1 for four in #h1hee1 for five in #cc1fdi1 for six in #ajj1idg1 for seven in #bde1aeg1 for eight in #dij1cad1 and for nine candidates in #hag1aij different ways4 ,add these together1 and we get the total of #a1fih1aei ways of voting4 #ajf4--,,the ,,muddletown ,,election4 ,the numbers of votes polled respectively by the ,liberal1 the ,conservative1 the ,independent1 and the ,socialist were #a1eec1 #a1ece1 #a1djg1 and #igh ,all that was necessary was to add the sum of the three majorities 7#gci7 to the total poll of #e1dgc 7making #f1bab7 and divide by #d1 which gives us #a1eec as the poll of the ,liberal4 ,then the polls of the other three candidates can1 of course1 be found by deducting the successive majorities from the last-mentioned number4 #ajg4--,,the ,,suffragists' ,,meeting4 ,eighteen were present at the meeting and eleven left4 ,if twelve had gone1 two-thirds would have retired4 ,if only nine had gone1 the meeting would have lost half its members4 #ajh4--,,the ,,leap-,,year ,,ladies4 ,the correct and only answer is that #aa1faf ladies made proposals of marriage4 ,here are all the details1 which the reader can check for himself with the original statements4 ,of #aj1afd spinsters1 #h1jhe married bachelors1 #fbg married widowers1 #a1bba were declined by bachelors1 and #bca declined by widowers4 ,of the #a1deb widows1 #a1aee married bachelors1 and #big married widowers4 ,no widows were declined4 ,the problem is not difficult1 by algebra1 when once we have succeeded in correctly stating it4 #aji4--,,the ,,great ,,scramble4 ,the smallest number of sugar plums that will fulfil the conditions is #bf1hhj4 ,the five boys obtained respectively3 ,andrew1 #b1hfc2 ,bob1 #f1cce2 ,charlie1 #b1dch2 ,david1 #aj1bid2 ,edgar1 #d1iej4 ,there is a little trap concealed in the words near the end1 one-fifth of the same1 that seems at first sight to upset the whole account of the affair4 ,but a little thought will show that the words could only mean one-fifth of five-eighths1 the fraction last mentioned--that is1 one-eighth of the three-quarters that ,bob and ,andrew had last acquired4 #aaj4--,,the ,,abbot',s ,,puzzle4 ,the only answer is that there were #e men1 #be women1 and #gj children4 ,there were thus #ajj persons in all1 #e times as many women as men1 and as the men would together receive #ae bushels1 the women #ej bushels1 and the children #ce bushels1 exactly #ajj bushels would be distributed4 #aaa4--,,reaping ,,the ,,corn4 ,the whole field must have contained #df4fbf square rods4 ,the side of the central square1 left by the farmer1 is #d4hbhd rods1 so it contains #bc4cac square rods4 ,the area of the field was thus something more than a quarter of an acre and less than one-third2 to be more precise1 4#biad of an acre4 #aab4--;,a ,,puzzling ,,legacy4 ,as the share of ,charles falls in through his death1 we have merely to divide the whole hundred acres between ,alfred and ,benjamin in the proportion of one-third to one-fourth--that is in the proportion of four-twelfths to three-twelfths1 which is the same as four to three4 ,therefore ,alfred takes four-sevenths of the hundred acres and ,benjamin three-sevenths4 #aac4--,,the ,,torn ,,number4 ,the other number that answers all the requirements of the puzzle is #i1hja4 ,if we divide this in the middle into two numbers and add them together we get #ii1 which1 multiplied by itself1 produces #i1hja4 ,it is true that #b1jbe may be treated in the same way1 only this number is excluded by the condition which requires that no two figures should be alike4 ,the general solution is curious4 ,call the number of figures in each half of the torn label n4 ,then1 if we add #a to each of the exponents of the prime factors 7other than #c7 of #ajn - #a 7#a being regarded as a factor with the constant exponent1 #a71 their product will be the number of solutions4 ,thus1 for a label of six figures1 n #c4 ,the factors of #ajn - #a are #a #cg 7not considering the #c71 and the product of #b #b #d1 the number of solutions4 ,this always includes the special cases #ih - #ja1 #jj - #ja1 #iih - #ja1 #jjj - #jja1 etc4 ,the solutions are obtained as follows3--,factorize #aj - #a in all possible ways1 always keeping the powers of #c together1 thus1 #cg #bg1 #iii #a4 ,then solve the equation #cgx #bgy #a4 ,here x #ai and y #bf4 ,therefore1 #ai #cg #gjc1 the square of which gives one label1 #did1bji4 ,a complementary solution 7through #bgx #cgx #a7 can at once be found by #ajn - #gjc #big1 the square of which gives #jhh1bji for second label4 7,these non-significant noughts to the left must be included1 though they lead to peculiar cases like #jjbch - #jdfda #dhgi1 where #jbch - #dfda would not work47 ,the special case #iii #a we can write at once #iih1jja1 according to the law shown above1 by adding nines on one half and noughts on the other1 and its complementary will be #a preceded by five noughts1 or #jjjjja4 ,thus we get the squares of #iii and #a4 ,these are the four solutions4 #aad4--;,,curious ,,numbers4 ,the three smallest numbers1 in addition to #dh1 are #a1fhj1 #eg1abj1 and #a1idj1ddh4 ,it will be found that #a1fha and #hda1 #eg1aba and #bh1efa1 #a1idj1ddi and #igj1bbe1 are respectively the squares of #da and #bi1 #bci and #afi1 #a1cic and #ihe4 #aae4--;,a ,,printer',s ,,error4 ,the answer is that #be 4#ib is the same as #beib1 and this is the only possible solution to the puzzle4 #aaf4--,,the ,,converted ,,miser4 ,as we are not told in what year ,mr4 ,jasper ,bullyon made the generous distribution of his accumulated wealth1 but are required to find the lowest possible amount of money1 it is clear that we must look for a year of the most favourable form4 ,there are four cases to be considered--an ordinary year with fifty-two ,sundays and with fifty-three ,sundays1 and a leap-year with fifty-two and fifty-three ,sundays respectively4 ,here are the lowest possible amounts in each case3-- #cac weekdays1 #eb ,sundays #aab1jee #cab weekdays1 #ec ,sundays #ai1cde #cad weekdays1 #eb ,sundays ,no solution possible4 #cac weekdays1 #ec ,sundays #fi1agd ,the lowest possible amount1 and therefore the correct answer1 is #ai1cde1 distributed in an ordinary year that began on a ,sunday4 ,the last year of this kind was #aiaa4 ,he would have paid #ec on every day of the year1 or #fb on every weekday1 with #a left over1 as required1 in the latter event4 #aag4--;,a ,,fence ,,problem4 ,though this puzzle presents no great difficulty to any one possessing a knowledge of algebra1 it has perhaps rather interesting features4 ,seeing1 as one does in the illustration1 just one corner of the proposed square1 one is scarcely prepared for the fact that the field1 in order to comply with the conditions1 must contain exactly #eja1gfj acres1 the fence requiring the same number of rails4 ,yet this is the correct answer1 and the only answer1 and if that gentleman in ,iowa carries out his intention1 his field will be twenty-eight miles long on each side1 and a little larger than the county of ,westmorland4 ,i am not aware that any limit has ever been fixed to the size of a field1 though they do not run so large as this in ,great ,britain4 ,still1 out in ,iowa1 where my correspondent resides1 they do these things on a very big scale4 ,i have1 however1 reason to believe that when he finds the sort of task he has set himself1 he will decide to abandon it2 for if that cow decides to roam to fresh woods and pastures new1 the milkmaid may have to start out a week in advance in order to obtain the morning's milk4 ,here is a little rule that will always apply where the length of the rail is half a pole4 ,multiply the number of rails in a hurdle by four1 and the result is the exact number of miles in the side of a square field containing the same number of acres as there are rails in the complete fence4 ,thus1 with a one-rail fence the field is four miles square2 a two-rail fence gives eight miles square2 a three-rail fence1 twelve miles square2 and so on1 until we find that a seven-rail fence multiplied by four gives a field of twenty-eight miles square4 ,in the case of our present problem1 if the field be made smaller1 then the number of rails will exceed the number of acres2 while if the field be made larger1 the number of rails will be less than the acres of the field4 #aah4--;,,circling ,,the ,,squares4 ,though this problem might strike the novice as being rather difficult1 it is1 as a matter of fact1 quite easy1 and is made still easier by inserting four out of the ten numbers4 ,first1 it will be found that squares that are diametrically opposite have a common difference4 ,for example1 the difference between the square of #ad and the square of #b1 in the diagram1 is #aib2 and the difference between the square of #af and the square of #h is also #aib4 ,this must be so in every case4 ,then it should be remembered that the difference between squares of two consecutive numbers is always twice the smaller number plus #a1 and that the difference between the squares of any two numbers can always be expressed as the difference of the numbers multiplied by their sum4 ,thus the square of #e 7#be7 less the square of #d 7#af7 equals 7#b #d7 #a1 or #i2 also1 the square of #g 7#di7 less the square of #c 7#i7 equals 7#g #c7 7#g - #c71 or #dj4 ,now1 the number #aib1 referred to above1 may be divided into five different pairs of even factors3 #b #if1 #d #dh1 #f #cb1 #h #bd1 and #ab #af1 and these divided by #b give us1 #a #dh1 #b #bd1 #c #af1 #d #ab1 and #f #h4 ,the difference and sum respectively of each of these pairs in turn produce #dg1 #di2 #bb1 #bf2 #ac1 #ai2 #h1 #af2 and #b1 #ad4 ,these are the required numbers1 four of which are already placed4 ,the six numbers that have to be added may be placed in just six different ways1 one of which is as follows1 reading round the circle clockwise3 #af1 #b1 #di1 #bb1 #ai1 #h1 #ad1 #dg1 #bf1 #ac4 ,i will just draw the reader's attention to one other little point4 ,in all circles of this kind1 the difference between diametrically opposite numbers increases by a certain ratio1 the first numbers 7with the exception of a circle of #f7 being #d and #f1 and the others formed by doubling the next preceding but one4 ,thus1 in the above case1 the first difference is #b1 and then the numbers increase by #d1 #f1 #h1 and #ab4 ,of course1 an infinite number of solutions may be found if we admit fractions4 ,the number of squares in a circle of this kind must1 however1 be of the form #dn #f2 that is1 it must be a number composed of #f plus a multiple of #d4 #aai4--,,rackbrane',s ,,little ,,loss4 ,the professor must have started the game with thirteen shillings1 ,mr4 ,potts with four shillings1 and ,mrs4 ,potts with seven shillings4 #abj4--,,the ,,farmer ,,and ,,his ,,sheep4 ,the farmer had one sheep only6 ,if he divided this sheep 7which is best done by weight7 into two parts1 making one part two-thirds and the other part one-third1 then the difference between these two numbers is the same as the difference between their squares--that is1 one-third4 ,any two fractions will do if the denominator equals the sum of the two numerators4 #aba4--;,,heads ,,or ,,tails4 ,crooks must have lost1 and the longer he went on the more he would lose4 ,in two tosses he would be left with three-quarters of his money1 in four tosses with nine-sixteenths of his money1 in six tosses with twenty-seven sixty-fourths of his money1 and so on4 ,the order of the wins and losses makes no difference1 so long as their number is in the end equal4 #abb4--,,the ,,see-,,saw ,,puzzle4 ,the boy's weight must have been about #ci4gi lbs4 ,a brick weighed #c lbs4 ,therefore #af bricks weighed #dh lbs4 and #aa bricks #cc lbs4 ,multiply #dh by #cc and take the square root4 #abc4--;,a ,,legal ,,difficulty4 ,it was clearly the intention of the deceased to give the son twice as much as the mother1 or the daughter half as much as the mother4 ,therefore the most equitable division would be that the mother should take two-sevenths1 the son four-sevenths1 and the daughter one-seventh4 #abd4--;,a ,,question ,,of ,,definition4 ,there is1 of course1 no difference in area between a mile square and a square mile4 ,but there may be considerable difference in shape4 ,a mile square can be no other shape than square2 the expression describes a surface of a certain specific size and shape4 ,a square mile may be of any shape2 the expression names a unit of area1 but does not prescribe any particular shape4 #abe4--,,the ,,miners' ,,holiday4 ,bill ,harris must have spent thirteen shillings and sixpence1 which would be three shillings more than the average for the seven men--half a guinea4 #abf4--,,simple ,,multiplication4 ,the number required is #c1ebi1daa1gfd1gje1hhb1 which may be multiplied by #c and divided by #b1 by the simple expedient of removing the #c from one end of the row to the other4 ,if you want a longer number1 you can increase this one to any extent by repeating the sixteen figures in the same order4 #abg4--,,simple ,,division4 ,subtract every number in turn from every other number1 and we get #ceh 7twice71 #gaf1 #a1faa1 #a1bec1 and #hie4 ,now1 we see at a glance that1 as #ceh equals #b #agi1 the only number that can divide in every case without a remainder will be #agi4 ,on trial we find that this is such a divisor4 ,therefore1 #agi is the divisor we want1 which always leaves a remainder #afd in the case of the original numbers given4 #abh4--;,a ,,problem ,,in ,,squares4 ,the sides of the three boards measure #ca in41 #da in41 and #di in4 ,the common difference of area is exactly five square feet4 ,three numbers whose squares are in ,a4,p41 with a common difference of #g1 are #aac_/abj1 #ccg_/abj1 #dfc_/abj2 and with a common difference of #ac are #hjibi_/aichj1 #ajfiba_/aichj1 and #abggbi_/aichj4 ,in the case of whole square numbers the common difference will always be divisible by #bd1 so it is obvious that our squares must be fractional4 ,readers should now try to solve the case where the common difference is #bc4 ,it is rather a hard nut4 #abi4--,,the ,,battle ,,of ,,hastings4 ,any number 7not itself a square number7 may be multiplied by a square that will give a product #a less than another square4 ,the given number must not itself be a square1 because a square multiplied by a square produces a square1 and no square plus #a can be a square4 ,my remarks throughout must be understood to apply to whole numbers1 because fractional soldiers are not of much use in war4 ,now1 of all the numbers from #b to #ii inclusive1 #fa happens to be the most awkward one to work1 and the lowest possible answer to our puzzle is that ,harold's army consisted of #c1aai1hhb1ihb1hfj1bfd1djj men4 ,that is1 there would be #ea1ade1fbb1ffi1hdj1djj men 7the square of #bbf1aec1ihj7 in each of the sixty-one squares4 ,add one man 7,harold71 and they could then form one large square with #a1gff1cai1jdi men on every side4 ,the general problem1 of which this is a particular case1 is known as the ,pellian ,equation--apparently because ,pell neither first propounded the question nor first solved it6 ,it was issued as a challenge by ,fermat to the ,english mathematicians of his day4 ,it is readily solved by the use of continued fractions4 ,next to #fa1 the most difficult number under #ajj is #ig1 where #ig #f1cgg1ceb #a a square4 ,the reason why ,i assumed that there must be something wrong with the figures in the chronicle is that we can confidently say that ,harold's army did not contain over three trillion men6 ,if this army 7not to mention the ,normans7 had had the whole surface of the earth 7sea included7 on which to encamp1 each man would have had slightly more than a quarter of a square inch of space in which to move about6 ,put another way3 ,allowing one square foot of standing-room per man1 each small square would have required all the space allowed by a globe three times the diameter of the earth4 #acj4--,,the ,,sculptor',s ,,problem4 ,a little thought will make it clear that the answer must be fractional1 and that in one case the numerator will be greater and in the other case less than the denominator4 ,as a matter of fact1 the height of the larger cube must be #h_/g ft41 and of the smaller #c_/g ft41 if we are to have the answer in the smallest possible figures4 ,here the lineal measurement is #aa_/g ft4--that is1 #ad_/g ft4 ,what are the cubic contents of the two cubes8 ,first #h_/g #c_/g #h_/g #eab_/cdc1 and secondly #c_/g #c_/g #c_/g #bg_/cdc4 ,add these together and the result is #eci_/cdc1 which reduces to #aa_/g or #ad_/g ft4 ,we thus see that the answers in cubic feet and lineal feet are precisely the same4 ,the germ of the idea is to be found in the works of ,diophantus of ,alexandria1 who wrote about the beginning of the fourth century4 ,these fractional numbers appear in triads1 and are obtained from three generators1 a1 b1 c1 where a is the largest and c the smallest4 ,then ab c denominator1 and a - c1 b - c1 and a - b will be the three numerators4 ,thus1 using the generators #c1 #b1 #a1 we get #h_/g1 #c_/g1 #e_/g and we can pair the first and second1 as in the above solution1 or the first and third for a second solution4 ,the denominator must always be a prime number of the form #fn #a1 or composed of such primes4 ,thus you can have #ac1 #ai1 etc41 as denominators1 but not #be1 #ee1 #ahg1 etc4 ,when the principle is understood there is no difficulty in writing down the dimensions of as many sets of cubes as the most exacting collector may require4 ,if the reader would like one1 for example1 with plenty of nines1 perhaps the following would satisfy him3 #iiiiiiii_/iiiijjja and #aiiii_/iiiijjja4 #aca4--,,the ,,spanish ,,miser4 ,there must have been #chf doubloons in one box1 #h1dej in another1 and #af1ead in the third1 because #chf is the smallest number that can occur4 ,if ,i had asked for the smallest aggregate number of coins1 the answer would have been #dhb1 #c1cfb1 and #f1bdb4 ,it will be found in either case that if the contents of any two of the three boxes be combined1 they form a square number of coins4 ,it is a curious coincidence 7nothing more1 for it will not always happen7 that in the first solution the digits of the three numbers add to #ag in every case1 and in the second solution to #ad4 ,it should be noted that the middle one of the three numbers will always be half a square4 #acb4--,,the ,,nine ,,treasure ,,boxes4 ,here is the answer that fulfils the conditions3-- ,a #d ,b #c1cfd ,c #f1gbd ,d #b1aaf ,e #e1dgf ,f #h1hcf ,g #i1dji ,h #ab1gfi ,i #af1abi ,each of these is a square number1 the roots1 taken in alphabetical order1 being #b1 #eh1 #hb1 #df1 #gd1 #id1 #ig1 #aac1 and #abg1 while the required difference between ,a and ,b1 ,b and ,c1 ,d and ,e4 etc41 is in every case #c1cfj4 #acc4--,,the ,,five ,,brigands4 ,the sum of #bjj doubloons might have been held by the five brigands in any one of #f1fbg different ways4 ,alfonso may have held any number from #a to #aa4 ,if he held #a doubloon1 there are #a1jje different ways of distributing the remainder2 if he held #b1 there are #ihe ways2 if #c1 there are #igg ways2 if #d1 there are #ijc ways2 if #e doubloons1 #hcb ways2 if #f doubloons1 #gjd ways2 if #g doubloons1 #egj ways2 if #h doubloons1 #chh ways2 if #i doubloons1 #bjj ways2 if #aj doubloons1 #fj ways2 and if ,alfonso held #aa doubloons1 the remainder could be distributed in #c different ways4 ,more than #aa doubloons he could not possibly have had4 ,it will scarcely be expected that ,i shall give all these #f1fbg ways at length4 ,what ,i propose to do is to enable the reader1 if he should feel so disposed1 to write out all the answers where ,alfonso has one and the same amount4 ,let us take the cases where ,alfonso has #f doubloons1 and see how we may obtain all the #gjd different ways indicated above4 ,here are two tables that will serve as keys to all these answers3-- ,table ,i4 ,table ,,ii4 ,a #f4 ,a #f4 ,b n4 ,b n4 ,c 7#fc - #en7 m4 ,c #a m4 ,d 7#abh #dn7 - #dm4 ,d 7#cgf - #afn7 - #dm4 ,e #c #cm4 ,e 7#aen - #ahc7 #cm4 ,in the first table we may substitute for n any whole number from #a to #ab inclusive1 and m may be nought or any whole number from #a to 7#ca n7 inclusive4 ,in the second table n may have the value of any whole number from #ac to #bc inclusive1 and m may be nought or any whole number from #a to 7#ic - #dn7 inclusive4 ,the first table thus gives 7#cb n7 answers for every value of n2 and the second table gives 7#id - #dn7 answers for every value of n4 ,the former1 therefore1 produces #dfb and the latter #bdb answers1 which together make #gjd1 as already stated4 ,let us take ,table ,i41 and say n #e and m #b2 also in ,table ,,ii4 take n #ac and m #j4 ,then we at once get these two answers3-- ,a #f ,a #f ,b #e ,b #ac ,c #dj ,c #a ,d #adj ,d #afh ,e #i ,e #ab --- --- #bjj doubloons #bjj doubloons4 ,these will be found to work correctly4 ,all the rest of the #gjd answers1 where ,alfonso always holds six doubloons1 may be obtained in this way from the two tables by substituting the different numbers for the letters m and n4 ,put in another way1 for every holding of ,alfonso the number of answers is the sum of two arithmetical progressions1 the common difference in one case being #a and in the other -#d4 ,thus in the case where ,alfonso holds #f doubloons one progression is #cc #cd #ce #cf 444 #dc #dd1 and the other #db #ch #cd #cj 444 #f #b4 ,the sum of the first series is #dfb1 and of the second #bdb--results which again agree with the figures already given4 ,the problem may be said to consist in finding the first and last terms of these progressions4 ,i should remark that where ,alfonso holds #i1 #aj1 or #aa there is only one progression1 of the second form4 #acd4--,,the ,,banker',s ,,puzzle4 ,in order that a number of sixpences may not be divisible into a number of equal piles1 it is necessary that the number should be a prime4 ,if the banker can bring about a prime number1 he will win2 and ,i will show how he can always do this1 whatever the customer may put in the box1 and that therefore the banker will win to a certainty4 ,the banker must first deposit forty sixpences1 and then1 no matter how many the customer may add1 he will desire the latter to transfer from the counter the square of the number next below what the customer put in4 ,thus1 banker puts #dj1 customer1 we will say1 adds #f1 then transfers from the counter #be 7the square of #e71 which leaves #ga in all1 a prime number4 ,try again4 ,banker puts #dj1 customer adds #ab1 then transfers #aba 7the square of #aa71 as desired1 which leaves #agc1 a prime number4 ,the key to the puzzle is the curious fact that any number up to #ci1 if added to its square and the sum increased by #da1 makes a prime number4 ,this was first discovered by ,euler1 the great mathematician4 ,it has been suggested that the banker might desire the customer to transfer sufficient to raise the contents of the box to a given number2 but this would not only make the thing an absurdity1 but breaks the rule that neither knows what the other puts in4 #ace4--,,the ,,stonemason',s ,,problem4 ,the puzzle amounts to this4 ,find the smallest square number that may be expressed as the sum of more than three consecutive cubes1 the cube #a being barred4 ,as more than three heaps were to be supplied1 this condition shuts out the otherwise smallest answer1 #bc #bd #be #bjd4 ,but it admits the answer1 #be #bf #bg #bh #bi #cae4 ,the correct answer1 however1 requires more heaps1 but a smaller aggregate number of blocks4 ,here it is3 #ad #ae 444 up to #be inclusive1 or twelve heaps in all1 which1 added together1 make #ig1cdd blocks of stone that may be laid out to form a square #cab #cab4 ,i will just remark that one key to the solution lies in what are called triangular numbers4 7,see pp4 #ac1 #be1 and #aff47 #acf4--,,the ,,sultan',s ,,army4 ,the smallest primes of the form #dn #a are #e1 #ac1 #ag1 #bi1 and #cg1 and the smallest of the form #dn - #a are #c1 #g1 #aa1 #ai1 and #bc4 ,now1 primes of the first form can always be expressed as the sum of two squares1 and in only one way4 ,thus1 #e #d #a2 #ac #i #d2 #ag #af #a2 #bi #be #d2 #cg #cf #a4 ,but primes of the second form can never be expressed as the sum of two squares in any way whatever4 ,in order that a number may be expressed as the sum of two squares in several different ways1 it is necessary that it shall be a composite number containing a certain number of primes of our first form4 ,thus1 #e or #ac alone can only be so expressed in one way2 but #fe1 7#e #ac71 can be expressed in two ways1 #a1aje1 7#e #ac #ag71 in four ways1 #cb1jde1 7#e #ac #ag #bi71 in eight ways4 ,we thus get double as many ways for every new factor of this form that we introduce4 ,note1 however1 that ,i say new factor1 for the repetition of factors is subject to another law4 ,we cannot express #be1 7#e #e71 in two ways1 but only in one2 yet #abe1 7#e #e #e71 can be given in two ways1 and so can #fbe1 7#e #e #e #e72 while if we take in yet another #e we can express the number as the sum of two squares in three different ways4 ,if a prime of the second form gets into your composite number1 then that number cannot be the sum of two squares4 ,thus #ae1 7#c #e71 will not work1 nor will #ace1 7#c #c #c #e72 but if we take in an even number of #c's it will work1 because these #c's will themselves form a square number1 but you will only get one solution4 ,thus1 #de1 7#c #c #e1 or #i #e7 #cf #i4 ,similarly1 the factor #b may always occur1 or any power of #b1 such as #d1 #h1 #af1 #cb2 but its introduction or omission will never affect the number of your solutions1 except in such a case as #ej1 where it doubles a square and therefore gives you the two answers1 #di #a and #be #be4 ,now1 directly a number is decomposed into its prime factors1 it is possible to tell at a glance whether or not it can be split into two squares2 and if it can be1 the process of discovery in how many ways is so simple that it can be done in the head without any effort4 ,the number ,i gave was #acj4 ,i at once saw that this was #b #e #ac1 and consequently that1 as #fe can be expressed in two ways 7#fd #a and #di #af71 #acj can also be expressed in two ways1 the factor #b not affecting the question4 ,the smallest number that can be expressed as the sum of two squares in twelve different ways is #afj1bbe1 and this is therefore the smallest army that would answer the ,sultan's purpose4 ,the number is composed of the factors #e #e #ac #ag #bi1 each of which is of the required form4 ,if they were all different factors1 there would be sixteen ways2 but as one of the factors is repeated1 there are just twelve ways4 ,here are the sides of the twelve pairs of squares3 7#djj and #ae71 7#cii and #cb71 7#cic and #gf71 7#cib and #ha71 7#chd and #aac71 7#cge and #adj71 7#cfj and #age71 7#cef and #ahc71 7#ccg and #baf71 7#cbi and #bbh71 7#caa and #beb71 7#bfe and #cjj74 ,square the two numbers in each pair1 add them together1 and their sum will in every case be #afj1bbe4 #acg4--;,a ,,study ,,in ,,thrift4 ,mrs4 ,sandy ,mc,allister will have to save a tremendous sum out of her housekeeping allowance if she is to win that sixth present that her canny husband promised her4 ,and the allowance must be a very liberal one if it is to admit of such savings4 ,the problem required that we should find five numbers higher than #cf the units of which may be displayed so as to form a square1 a triangle1 two triangles1 and three triangles1 using the complete number in every one of the four cases4 ,every triangular number is such that if we multiply it by #h and add #a the result is an odd square number4 ,for example1 multiply #a1 #c1 #f1 #aj1 #ae respectively by #h and add #a1 and we get #i1 #be1 #di1 #ha1 #aba1 which are the squares of the odd numbers #c1 #e1 #g1 #i1 #aa4 ,therefore in every case where #hx #a a square number1 x is also a triangular4 ,this point is dealt with in our puzzle1 ,the ,battle of ,hastings4 ,i will now merely show again how1 when the first solution is found1 the others may be discovered without any difficulty4 ,first of all1 here are the figures3-- #h #a #a #c #h #f #a #ag #h #ce #a #ii #h #bjd #a #egg #h #aahi #a #ccfc #h #ficj #a #aifja #h #djcia #a #aadbdc ,the successive pairs of numbers are found in this way3-- 7#a #c7 7#c #a7 #f 7#h #a7 7#c #c7 #ag 7#a #ag7 7#c #f7 #ce 7#h #f7 7#c #ag7 #ii 7#a #ii7 7#c #ce7 #bjd 7#h #ce7 7#c #ii7 #egg and so on4 ,look for the numbers in the table above1 and the method will explain itself4 ,thus we find that the numbers #cf1 #abbe1 #dafaf1 #adacgba1 #dhjbdijj1 and #afcadcbhha will form squares with sides of #f1 #ce1 #bjd1 #aahi1 #ficj1 and #djcia2 and they will also form single triangles with sides of #h1 #di1 #bhh1 #afha1 #ihjj1 and #egaba4 ,these numbers may be obtained from the last column in the first table above in this way3 simply divide the numbers by #b and reject the remainder4 ,thus the integral halves of #ag1 #ii1 and #egg are #h1 #di1 and #bhh4 ,all the numbers we have found will form either two or three triangles at will4 ,the following little diagram will show you graphically at a glance that every square number must necessarily be the sum of two triangulars1 and that the side of one triangle will be the same as the side of the corresponding square1 while the other will be just #a less4 ,illustration ----------- --------- 4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4_/4 4 4 4 4 _/4 4 4 4 4 _/4 4 4 4 4 4 --------- ----------- ,thus a square may always be divided easily into two triangles1 and the sum of two consecutive triangulars will always make a square4 ,in numbers it is equally clear1 for if we examine the first triangulars--#a1 #c1 #f1 #aj1 #ae1 #ba1 #bh--we find that by adding all the consecutive pairs in turn we get the series of square numbers--#i1 #af1 #be1 #cf1 #di1 etc4 ,the method of forming three triangles from our numbers is equally direct1 and not at all a matter of trial4 ,but ,i must content myself with giving actual figures1 and just stating that every triangular higher than #f will form three triangulars4 ,i give the sides of the triangles1 and readers will know from my remarks when stating the puzzle how to find from these sides the number of counters or coins in each1 and so check the results if they so wish4 ---------------------------------------- ------------------------------- ,number ,side of ,side of ,sides of ,two ,sides of ,three ,square4 ,triangle4 ,triangles4 ,triangles4 ---------------------------------------- ------------------------------ #cf #f #h #f #e #e #e #c #abbe #ce #di #cf #cd #cc #cb #af #dafaf #bjd #bhh #bjd #bjc #aib #aib #ie #adacgba #aahi #afha #aahi #aahh #aaba #aabj #efj #dhjbdijj #ficj #ihjj #ficj #fibi #fecc #fecc #cbfg #afcadcbhha #djcia #egaba #djcia #djcij #chjha #chjhj #aijdj ---------------------------------------- ------------------------------ ,i should perhaps explain that the arrangements given in the last two columns are not the only ways of forming two and three triangles4 ,there are others1 but one set of figures will fully serve our purpose4 ,we thus see that before ,mrs4 ,mc,allister can claim her sixth #e present she must save the respectable sum of #a1fca1dcb1hha4 #ach4--,,the ,,artillerymen',s ,,dilemma4 ,we were required to find the smallest number of cannon balls that we could lay on the ground to form a perfect square1 and could pile into a square pyramid4 ,i will try to make the matter clear to the merest novice4 #a #b #c #d #e #f #g #a #c #f #aj #ae #ba #bh #a #d #aj #bj #ce #ef #hd #a #e #ad #cj #ee #ia #adj ,here in the first row we place in regular order the natural numbers4 ,each number in the second row represents the sum of the numbers in the row above1 from the beginning to the number just over it4 ,thus #a1 #b1 #c1 #d1 added together1 make #aj4 ,the third row is formed in exactly the same way as the second4 ,in the fourth row every number is formed by adding together the number just above it and the preceding number4 ,thus #d and #aj make #ad1 #bj and #ce make #ee4 ,now1 all the numbers in the second row are triangular numbers1 which means that these numbers of cannon balls may be laid out on the ground so as to form equilateral triangles4 ,the numbers in the third row will all form our triangular pyramids1 while the numbers in the fourth row will all form square pyramids4 ,thus the very process of forming the above numbers shows us that every square pyramid is the sum of two triangular pyramids1 one of which has the same number of balls in the side at the base1 and the other one ball fewer4 ,if we continue the above table to twenty-four places1 we shall reach the number #d1ijj in the fourth row4 ,as this number is the square of #gj1 we can lay out the balls in a square1 and can form a square pyramid with them4 ,this manner of writing out the series until we come to a square number does not appeal to the mathematical mind1 but it serves to show how the answer to the particular puzzle may be easily arrived at by anybody4 ,as a matter of fact1 ,i confess my failure to discover any number other than #d1ijj that fulfils the conditions1 nor have ,i found any rigid proof that this is the only answer4 ,the problem is a difficult one1 and the second answer1 if it exists 7which ,i do not believe71 certainly runs into big figures4 ,for the benefit of more advanced mathematicians ,i will add that the general expression for square pyramid numbers is 7#bn #cn n7_/#f4 ,for this expression to be also a square number 7the special case of #a excepted7 it is necessary that n p - #a #ft1 where #bp - #a q 7the ,pellian ,equation74 ,in the case of our solution above1 n #bd1 p #e1 t #b1 q #g4 #aci4--,,the ,,dutchmen',s ,,wives4 ,the money paid in every case was a square number of shillings1 because they bought #a at #as41 #b at #bs41 #c at #cs41 and so on4 ,but every husband pays altogether #fcs4 more than his wife1 so we have to find in how many ways #fc may be the difference between two square numbers4 ,these are the three only possible ways3 the square of #h less the square of #a1 the square of #ab less the square of #i1 and the square of #cb less the square of #ca4 ,here #a1 #i1 and #ca represent the number of pigs bought and the number of shillings per pig paid by each woman1 and #h1 #ab1 and #cb the same in the case of their respective husbands4 ,from the further information given as to their purchases1 we can now pair them off as follows3 ,cornelius and ,gurtrn bought #h and #a2 ,elas and ,katrn bought #ab and #i2 ,hendrick and ,anna bought #cb and #ca4 ,and these pairs represent correctly the three married couples4 ,the reader may here desire to know how we may determine the maximum number of ways in which a number may be expressed as the difference between two squares1 and how we are to find the actual squares4 ,any integer except #a1 #d1 and twice any odd number1 may be expressed as the difference of two integral squares in as many ways as it can be split up into pairs of factors1 counting #a as a factor4 ,suppose the number to be #e1idj4 ,the factors are #b4c4e4aa4 ,here the exponents are #b1 #c1 #a1 #a4 ,always deduct #a from the exponents of #b and add #a to all the other exponents2 then we get #a1 #d1 #b1 #b1 and half the product of these four numbers will be the required number of ways in which #e1idj may be the difference of two squares--that is1 #h4 ,to find these eight squares1 as it is an even number1 we first divide by #d and get #adhe1 the eight pairs of factors of which are #a #adhe1 #c #die1 #e #big1 #i #afe1 #aa #ace1 #ae #ii1 #bg #ee1 and #cc #de4 ,the sum and difference of any one of these pairs will give the required numbers4 ,thus1 the square of #a1dhf less the square of #a1dhd is #e1idj1 the square of #dih less the square of #dib is the same1 and so on4 ,in the case of #fc above1 the number is odd2 so we factorize at once1 #a #fc1 #c #ba1 #g #i4 ,then we find that half the sum and difference will give us the numbers #cb and #ca1 #ab and #i1 and #h and #a1 as shown in the solution to the puzzle4 ,the reverse problem1 to find the factors of a number when you have expressed it as the difference of two squares1 is obvious4 ,for example1 the sum and difference of any pair of numbers in the last sentence will give us the factors of #fc4 ,every prime number 7except #a and #b7 may be expressed as the difference of two squares in one way1 and in one way only4 ,if a number can be expressed as the difference of two squares in more than one way1 it is composite2 and having so expressed it1 we may at once obtain the factors1 as we have seen4 ,fermat showed in a letter to ,mersenne or ,frnicle1 in #afdc1 how we may discover whether a number may be expressed as the difference of two squares in more than one way1 or proved to be a prime4 ,but the method1 when dealing with large numbers1 is necessarily tedious1 though in practice it may be considerably shortened4 ,in many cases it is the shortest method known for factorizing large numbers1 and ,i have always held the opinion that ,fermat used it in performing a certain feat in factorizing that is historical and wrapped in mystery4 #adj4--;,,find ,,ada',s ,,surname4 ,the girls' names were ,ada ,smith1 ,annie ,brown1 ,emily ,jones1 ,mary ,robinson1 and ,bessie ,evans4 #ada4--,,saturday ,,marketing4 ,as every person's purchase was of the value of an exact number of shillings1 and as the party possessed when they started out forty shilling coins altogether1 there was no necessity for any lady to have any smaller change1 or any evidence that they actually had such change4 ,this being so1 the only answer possible is that the women were named respectively ,anne ,jones1 ,mary ,robinson1 ,jane ,smith1 and ,kate ,brown4 ,it will now be found that there would be exactly eight shillings left1 which may be divided equally among the eight persons in coin without any change being required4 #adb4--,,the ,,silk ,,patchwork4 ,illustration ,our illustration will show how to cut the stitches of the patchwork so as to get the square ,f entire1 and four equal pieces1 ,g1 ,h1 ,i1 ,k1 that will form a perfect ,greek cross4 ,the reader will know how to assemble these four pieces from ,fig4 #ac in the article4 ,illustration3 ,fig4 #a4 ,illustration3 ,fig4 #b4 #adc4--,,two ,,crosses ,,from ,,one4 ,it will be seen that one cross is cut out entire1 as ,a in ,fig4 #a1 while the four pieces marked ,b1 ,c1 ,d and ,e form the second cross1 as in ,fig4 #b1 which will be of exactly the same size as the other4 ,i will leave the reader the pleasant task of discovering for himself the best way of finding the direction of the cuts4 ,note that the ,swastika again appears4 ,the difficult question now presents itself3 ,how are we to cut three ,greek crosses from one in the fewest possible pieces8 ,as a matter of fact1 this problem may be solved in as few as thirteen pieces2 but as ,i know many of my readers1 advanced geometricians1 will be glad to have something to work on of which they are not shown the solution1 ,i leave the mystery for the present undisclosed4 #add4--,,the ,,cross ,,and ,,the ,,triangle4 ,the line ,a ,b in the following diagram represents the side of a square having the same area as the cross4 ,i have shown elsewhere1 as stated1 how to make a square and equilateral triangle of equal area4 ,i need not go1 therefore1 into the preliminary question of finding the dimensions of the triangle that is to equal our cross4 ,we will assume that we have already found this1 and the question then becomes1 ,how are we to cut up one of these into pieces that will form the other8 ,first draw the line ,a ,b where ,a and ,b are midway between the extremities of the two side arms4 ,next make the lines ,d ,c and ,e ,f equal in length to half the side of the triangle4 ,now from ,e and ,f describe with the same radius the intersecting arcs at ,g and draw ,f ,g4 ,finally make ,i ,k equal to ,h ,c and ,l ,b equal to ,a ,d4 ,if we now draw ,i ,l1 it should be parallel to ,f ,g1 and all the six pieces are marked out4 ,these fit together and form a perfect equilateral triangle1 as shown in the second diagram4 ,or we might have first found the direction of the line ,m ,n in our triangle1 then placed the point ,o over the point ,e in the cross and turned round the triangle over the cross until the line ,m ,n was parallel to ,a ,b4 ,the piece #e can then be marked off and the other pieces in succession4 ,illustration ,i have seen many attempts at a solution involving the assumption that the height of the triangle is exactly the same as the height of the cross4 ,this is a fallacy3 the cross will always be higher than the triangle of equal area4 #ade4--,,the ,,folded ,,cross4 ,illustration3 ,,fig4 #a41 ,,fig #b4 ,first fold the cross along the dotted line ,a ,b in ,fig4 #a4 ,you then have it in the form shown in ,fig4 #b4 ,next fold it along the dotted line ,c ,d 7where ,d is1 of course1 the centre of the cross71 and you get the form shown in ,fig4 #c4 ,now take your scissors and cut from ,g to ,f1 and the four pieces1 all of the same size and shape1 will fit together and form a square1 as shown in ,fig4 #d4 ,illustration3 ,,fig4 #c41 ,,fig4 #d4 #adf4--;,,an ,,easy ,,dissection ,,puzzle4 ,illustration - 3 3 3 3 3 ----------- _/ 3 3 _/ 3 3 _/ 3 3 _/ 3 3 _/ 3 3 ,the solution to this puzzle is shown in the illustration4 ,divide the figure up into twelve equal triangles1 and it is easy to discover the directions of the cuts1 as indicated by the dark lines4 #adg4--;,,an ,,easy ,,square ,,puzzle4 ,illustration ---------------------------------------- - 4 _/ 4 _/ 4 _/ _/ _/ _/ 4 _/ _/ 4 _/ _/ 4 _/ _/ 4_/ -------------------- _/ _/ _/ _/ _/ _/ _/ 4 _/ _/ 4 _/ _/ 4 _/ _/ 4 _/ _/ 4 _/ 4 _/ 4 _/ 4 ---------------------------------------- - ,the diagram explains itself1 one of the five pieces having been cut in two to form a square4 #adh4--,,the ,,bun ,,puzzle4 ,illustration 4 4 4 4 4 4 4 ,a 4 4 ,c 4 _/ 4 _/ 4 4 4 ,b 4 4 4 4 4 - 4 4 4 4 4 4 ,d ,e 4 4 4 4 4 4 4 4 4 -- 4 4 4 4 4 - - 4 ,g ,f - - 4 4 4 4 4 - - 4 4 4 - -- 4 4 - - 4 ,h - - 4 4 - - ,the secret of the bun puzzle lies in the fact that1 with the relative dimensions of the circles as given1 the three diameters will form a right-angled triangle1 as shown by ,a1 ,b1 ,c4 ,it follows that the two smaller buns are exactly equal to the large bun4 ,therefore1 if we give ,david and ,edgar the two halves marked ,d and ,e1 they will have their fair shares--one quarter of the confectionery each4 ,then if we place the small bun1 ,h1 on the top of the remaining one and trace its circumference in the manner shown1 ,fred's piece1 ,f1 will exactly equal ,harry's small bun1 ,h1 with the addition of the piece marked ,g--half the rim of the other4 ,thus each boy gets an exactly equal share1 and there are only five pieces necessary4 #adi4--,,the ,,chocolate ,,squares4 ,illustration ,square ,a is left entire2 the two pieces marked ,b fit together and make a second square2 the two pieces ,c make a third square2 and the four pieces marked ,d will form the fourth square4 #aej4--;,,dissecting ,a ,,mitre4 ,the diagram on the next page shows how to cut into five pieces to form a square4 ,the dotted lines are intended to show how to find the points ,c and ,f--the only difficulty4 ,a ,b is half ,b ,d1 and ,a ,e is parallel to ,b ,h4 ,with the point of the compasses at ,b describe the arc ,h ,e1 and ,a ,e will be the distance of ,c from ,b4 ,then ,f ,g equals ,b ,c less ,a ,b4 ,this puzzle--with the added condition that it shall be cut into four parts of the same size and shape--,i have not been able to trace to an earlier date than #ahce4 ,strictly speaking1 it is1 in that form1 impossible of solution2 but ,i give the answer that is always presented1 and that seems to satisfy most people4 ,illustration ,we are asked to assume that the two portions containing the same letter--,,aa1 ,,bb1 ,,cc1 ,,dd--are joined by a mere hair1 and are1 therefore1 only one piece4 ,to the geometrician this is absurd1 and the four shares are not equal in area unless they consist of two pieces each4 ,if you make them equal in area1 they will not be exactly alike in shape4 ,illustration #aea4--,,the ,,joiner',s ,,problem4 ,illustration ,nothing could be easier than the solution of this puzzle--when you know how to do it4 ,and yet it is apt to perplex the novice a good deal if he wants to do it in the fewest possible pieces--three4 ,all you have to do is to find the point ,a1 midway between ,b and ,c1 and then cut from ,a to ,d and from ,a to ,e4 ,the three pieces then form a square in the manner shown4 ,of course1 the proportions of the original figure must be correct2 thus the triangle ,,bef is just a quarter of the square ,,bcdf4 ,draw lines from ,b to ,d and from ,c to ,f and this will be clear4 #aeb4--;,,another ,,joiner',s ,,problem4 ,illustration ,,the point was to find a general rule for forming a perfect square out of another square combined with a right-angled isosceles triangle4 ,the triangle to which geometricians give this high-sounding name is1 of course1 nothing more or less than half a square that has been divided from corner to corner4 ,the precise relative proportions of the square and triangle are of no consequence whatever4 ,it is only necessary to cut the wood or material into five pieces4 ,suppose our original square to be ,,aclf in the above diagram and our triangle to be the shaded portion ,,ced4 ,now1 we first find half the length of the long side of the triangle 7,,cd7 and measure off this length at ,,ab4 ,then we place the triangle in its present position against the square and make two cuts--one from ,b to ,f1 and the other from ,b to ,e4 ,strange as it may seem1 that is all that is necessary6 ,if we now remove the pieces ,g1 ,h1 and ,m to their new places1 as shown in the diagram1 we get the perfect square ,,bekf4 ,take any two square pieces of paper1 of different sizes but perfect squares1 and cut the smaller one in half from corner to corner4 ,now proceed in the manner shown1 and you will find that the two pieces may be combined to form a larger square by making these two simple cuts1 and that no piece will be required to be turned over4 ,the remark that the triangle might be a little larger or a good deal smaller in proportion was intended to bar cases where area of triangle is greater than area of square4 ,in such cases six pieces are necessary1 and if triangle and square are of equal area there is an obvious solution in three pieces1 by simply cutting the square in half diagonally4 #aec4--;,a ,,cutting-,,out ,,puzzle4 ,illustration ,the illustration shows how to cut the four pieces and form with them a square4 ,first find the side of the square 7the mean proportional between the length and height of the rectangle71 and the method is obvious4 ,if our strip is exactly in the proportions #i #a1 or #af #a1 or #be #a1 we can clearly cut it in #c1 #d1 or #e rectangular pieces respectively to form a square4 ,excluding these special cases1 the general law is that for a strip in length more than n times the breadth1 and not more than 7n#a7 times the breadth1 it may be cut in n#b pieces to form a square1 and there will be n-#a rectangular pieces like piece #d in the diagram4 ,thus1 for example1 with a strip #bd #a1 the length is more than #af and less than #be times the breadth4 ,therefore it can be done in #f pieces 7n here being #d71 #c of which will be rectangular4 ,in the case where n equals #a1 the rectangle disappears and we get a solution in three pieces4 ,within these limits1 of course1 the sides need not be rational3 the solution is purely geometrical4 #aed4--,,mrs4 ,,hobson',s ,,hearthrug4 ,illustration ,as ,i gave full measurements of the mutilated rug1 it was quite an easy matter to find the precise dimensions for the square4 ,the two pieces cut off would1 if placed together1 make an oblong piece #ab #f1 giving an area of #gb 7inches or yards1 as we please71 and as the original complete rug measured #cf #bg1 it had an area of #igb4 ,if1 therefore1 we deduct the pieces that have been cut away1 we find that our new rug will contain #igb less #gb1 or #ijj2 and as #ijj is the square of #cj1 we know that the new rug must measure #cj #cj to be a perfect square4 ,this is a great help towards the solution1 because we may safely conclude that the two horizontal sides measuring #cj each may be left intact4 ,there is a very easy way of solving the puzzle in four pieces1 and also a way in three pieces that can scarcely be called difficult1 but the correct answer is in only two pieces4 ,it will be seen that if1 after the cuts are made1 we insert the teeth of the piece ,b one tooth lower down1 the two portions will fit together and form a square4 #aee4--,,the ,,pentagon ,,and ,,square4 ,a regular pentagon may be cut into as few as six pieces that will fit together without any turning over and form a square1 as ,i shall show below4 ,hitherto the best answer has been in seven pieces--the solution produced some years ago by a foreign mathematician1 ,paul ,busschop4 ,we first form a parallelogram1 and from that the square4 ,the process will be seen in the diagram on the next page4 ,the pentagon is ,,abcde4 ,by the cut ,,ac and the cut ,,fm 7,f being the middle point between ,a and ,c1 and ,m being the same distance from ,a as ,f7 we get two pieces that may be placed in position at ,,ghea and form the parallelogram ,,ghdc4 ,we then find the mean proportional between the length ,,hd and the height of the parallelogram4 ,this distance we mark off from ,c at ,k1 then draw ,,ck1 and from ,g drop the line ,,gl1 perpendicular to ,,kc4 ,the rest is easy and rather obvious4 ,it will be seen that the six pieces will form either the pentagon or the square4 ,i have received what purported to be a solution in five pieces1 but the method was based on the rather subtle fallacy that half the diagonal plus half the side of a pentagon equals the side of a square of the same area4 ,i say subtle1 because it is an extremely close approximation that will deceive the eye1 and is quite difficult to prove inexact4 ,i am not aware that attention has before been drawn to this curious approximation4 ,illustration ,another correspondent made the side of his square #a of the side of the pentagon4 ,as a matter of fact1 the ratio is irrational4 ,i calculate that if the side of the pentagon is #a--;inch1 foot1 or anything else--the side of the square of equal area is #a4caag nearly1 or say roughly #ac_/aj4 ,so we can only hope to solve the puzzle by geometrical methods4 #aef4--,,the ,,dissected ,,triangle4 ,diagram ,a is our original triangle4 ,we will say it measures #e inches 7or #e feet7 on each side4 ,if we take off a slice at the bottom of any equilateral triangle by a cut parallel with the base1 the portion that remains will always be an equilateral triangle2 so we first cut off piece #a and get a triangle #c inches on every side4 ,the manner of finding directions of the other cuts in ,a is obvious from the diagram4 ,now1 if we want two triangles1 #a will be one of them1 and #b1 #c1 #d1 and #e will fit together1 as in ,b1 to form the other4 ,if we want three equilateral triangles1 #a will be one1 #d and #e will form the second1 as in ,c1 and #b and #c will form the third1 as in ,d4 ,in ,b and ,c the piece #e is turned over2 but there can be no objection to this1 as it is not forbidden1 and is in no way opposed to the nature of the puzzle4 ,illustration #aeg4--,,the ,,table-,,top ,,and ,,stools4 ,illustration ,one object that ,i had in view when presenting this little puzzle was to point out the uncertainty of the meaning conveyed by the word oval4 ,though originally derived from the ,latin word ovum1 an egg1 yet what we understand as the egg-shape 7with one end smaller than the other7 is only one of many forms of the oval2 while some eggs are spherical in shape1 and a sphere or circle is most certainly not an oval4 ,if we speak of an ellipse--a conical ellipse--we are on safer ground1 but here we must be careful of error4 ,i recollect a ,liverpool town councillor1 many years ago1 whose ignorance of the poultry-yard led him to substitute the word hen for fowl1 remarking1 ,we must remember1 gentlemen1 that although every cock is a hen1 every hen is not a cock6 ,similarly1 we must always note that although every ellipse is an oval1 every oval is not an ellipse4 ,it is correct to say that an oval is an oblong curvilinear figure1 having two unequal diameters1 and bounded by a curve line returning into itself2 and this includes the ellipse1 but all other figures which in any way approach towards the form of an oval without necessarily having the properties above described are included in the term oval4 ,thus the following solution that ,i give to our puzzle involves the pointed oval1 known among architects as the vesica piscis4 ,illustration3 ,,the ,,two ,,stools4 ,the dotted lines in the table are given for greater clearness1 the cuts being made along the other lines4 ,it will be seen that the eight pieces form two stools of exactly the same size and shape with similar hand-holes4 ,these holes are a trifle longer than those in the schoolmaster's stools1 but they are much narrower and of considerably smaller area4 ,of course #e and #f can be cut out in one piece--also #g and #h--making only six pieces in all4 ,but ,i wished to keep the same number as in the original story4 ,when ,i first gave the above puzzle in a ,london newspaper1 in competition1 no correct solution was received1 but an ingenious and neatly executed attempt by a man lying in a ,london infirmary was accompanied by the following note3 ,having no compasses here1 ,i was compelled to improvise a pair with the aid of a small penknife1 a bit of firewood from a bundle1 a piece of tin from a toy engine1 a tin tack1 and two portions of a hairpin1 for points4 ,they are a fairly serviceable pair of compasses1 and ,i shall keep them as a memento of your puzzle4 #aeh4--,,the ,,great ,,monad4 ,the areas of circles are to each other as the squares of their diameters4 ,if you have a circle #b in4 in diameter and another #d in4 in diameter1 then one circle will be four times as great in area as the other1 because the square of #d is four times as great as the square of #b4 ,now1 if we refer to ,diagram #a1 we see how two equal squares may be cut into four pieces that will form one larger square2 from which it is self-evident that any square has just half the area of the square of its diagonal4 ,in ,diagram #b ,i have introduced a square as it often occurs in ancient drawings of the ,monad2 which was my reason for believing that the symbol had mathematical meanings1 since it will be found to demonstrate the fact that the area of the outer ring or annulus is exactly equal to the area of the inner circle4 ,compare ,diagram #b with ,diagram #a1 and you will see that as the square of the diameter ,,cd is double the square of the diameter of the inner circle1 or ,,ce1 therefore the area of the larger circle is double the area of the smaller one1 and consequently the area of the annulus is exactly equal to that of the inner circle4 ,this answers our first question4 ,illustration3 #a #b #c #d ,in ,diagram #c ,i show the simple solution to the second question4 ,it is obviously correct1 and may be proved by the cutting and superposition of parts4 ,the dotted lines will also serve to make it evident4 ,the third question is solved by the cut ,,cd in ,diagram #b1 but it remains to be proved that the piece ,f is really one-half of the ,yin or the ,yan4 ,this we will do in ,diagram #d4 ,the circle ,k has one-quarter the area of the circle containing ,yin and ,yan1 because its diameter is just one-half the length4 ,also ,l in ,diagram #c is1 we know1 one-quarter the area4 ,it is therefore evident that ,g is exactly equal to ,h1 and therefore half ,g is equal to half ,h4 ,so that what ,f loses from ,l it gains from ,k1 and ,f must be half of ,yin or ,yan4 #aei4--,,the ,,square ,,of ,,veneer4 ,illustration3 --------------------------------------- 3 3 3 3 3 3 3 3 --------------------------------------- 3 3 3 3 3 3 3 3 --------------------------------------- 3 3 3 3 3333 --------------------------------------- 3 3 3 3 3 3 3 3 --------------------------------- 3 3 3 3 3 3 3 3 --------------------------------------- 3333 3 3 3 3 --------------------------------------- 3 3 3 3 3 3 ,b 3 3 ,c --------------------------------------- 3 3 3 3 3333 ------------------------ 3 3 3 3 3 3 3 3 --------------------------------------- 3 3 3 3 3 3 3 3 ,a --------------------------------------- 3333 3 3 3 3 --------------------------------------- 3 3 3 3 3 3 3 3 --------------- 3 3 3 3 3 ,d 3 3 3 ------------------------ ,any square number may be expressed as the sum of two squares in an infinite number of different ways4 ,the solution of the present puzzle forms a simple demonstration of this rule4 ,it is a condition that we give actual dimensions4 ,in this puzzle ,i ignore the known dimensions of our square and work on the assumption that it is #acn by #acn4 ,the value of n we can afterwards determine4 ,divide the square as shown 7where the dotted lines indicate the original markings7 into #afi squares4 ,as #afi is the sum of the two squares #add and #be1 we will proceed to divide the veneer into two squares1 measuring respectively #ab #ab and #e #e2 and as we know that two squares may be formed from one square by dissection in four pieces1 we seek a solution in this number4 ,the dark lines in the diagram show where the cuts are to be made4 ,the square #e #e is cut out whole1 and the larger square is formed from the remaining three pieces1 ,b1 ,c1 and ,d1 which the reader can easily fit together4 ,now1 n is clearly #e_/ac of an inch4 ,consequently our larger square must be #fj_/ac in4 #fj_/ac in41 and our smaller square #be_/ac in4 #be_/ac in4 ,the square of #fj_/ac added to the square of #be_/ac is #be4 ,the square is thus divided into as few as four pieces that form two squares of known dimensions1 and all the sixteen nails are avoided4 ,here is a general formula for finding two squares whose sum shall equal a given square1 say a4 ,in the case of the solution of our puzzle p #c1 q #b1 and a #e4 #bpqa _/ a7 p q7 - 7#bpqa7 --------- x2 --------------------------- y p q p q ,here x y a4 #afj4--,,the ,,two ,,horseshoes4 ,the puzzle was to cut the two shoes 7including the hoof contained within the outlines7 into four pieces1 two pieces each1 that would fit together and form a perfect circle4 ,it was also stipulated that all four pieces should be different in shape4 ,as a matter of fact1 it is a puzzle based on the principle contained in that curious ,chinese symbol the ,monad4 7,see ,no4 #aeh47 ,illustration ,the above diagrams give the correct solution to the problem4 ,it will be noticed that #a and #b are cut into the required four pieces1 all different in shape1 that fit together and form the perfect circle shown in ,diagram #c4 ,it will further be observed that the two pieces ,a and ,b of one shoe and the two pieces ,c and ,d of the other form two exactly similar halves of the circle--the ,yin and the ,yan of the great ,monad4 ,it will be seen that the shape of the horseshoe is more easily determined from the circle than the dimensions of the circle from the horseshoe1 though the latter presents no difficulty when you know that the curve of the long side of the shoe is part of the circumference of your circle4 ,the difference between ,b and ,d is instructive1 and the idea is useful in all such cases where it is a condition that the pieces must be different in shape4 ,in forming ,d we simply add on a symmetrical piece1 a curvilinear square1 to the piece ,b4 ,therefore1 in giving either ,b or ,d a quarter turn before placing in the new position1 a precisely similar effect must be produced4 #afa4--,,the ,,betsy ,,ross ,,puzzle4 ,fold the circular piece of paper in half along the dotted line shown in ,fig4 #a1 and divide the upper half into five equal parts as indicated4 ,now fold the paper along the lines1 and it will have the appearance shown in ,fig4 #b4 ,if you want a star like ,fig4 #c1 cut from ,a to ,b2 if you wish one like ,fig4 #d1 cut from ,a to ,c4 ,thus1 the nearer you cut to the point at the bottom the longer will be the points of the star1 and the farther off from the point that you cut the shorter will be the points of the star4 ,illustration #afb4--,,the ,,cardboard ,,chain4 ,the reader will probably feel rewarded for any care and patience that he may bestow on cutting out the cardboard chain4 ,we will suppose that he has a piece of cardboard measuring #h in4 by #b in41 though the dimensions are of no importance4 ,yet if you want a long chain you must1 of course1 take a long strip of cardboard4 ,first rule pencil lines ,b ,b and ,c ,c1 half an inch from the edges1 and also the short perpendicular lines half an inch apart4 7,see next page47 ,rule lines on the other side in just the same way1 and in order that they shall coincide it is well to prick through the card with a needle the points where the short lines end4 ,now take your penknife and split the card from ,a ,a down to ,b ,b1 and from ,d ,d up to ,c ,c4 ,then cut right through the card along all the short perpendicular lines1 and half through the card along the short portions of ,b ,b and ,c ,c that are not dotted4 ,next turn the card over and cut half through along the short lines on ,b ,b and ,c ,c at the places that are immediately beneath the dotted lines on the upper side4 ,with a little careful separation of the parts with the penknife1 the cardboard may now be divided into two interlacing ladder-like portions1 as shown in ,fig4 #b2 and if you cut away all the shaded parts you will get the chain1 cut solidly out of the cardboard1 without any join1 as shown in the illustrations on page #dj4 ,it is an interesting variant of the puzzle to cut out two keys on a ring--in the same manner without join4 ,illustration #afd4--,,the ,,potato ,,puzzle4 ,as many as twenty-two pieces may be obtained by the six cuts4 ,the illustration shows a pretty symmetrical solution4 ,the rule in such cases is that every cut shall intersect every other cut and no two intersections coincide2 that is to say1 every line passes through every other line1 but more than two lines do not cross at the same point anywhere4 ,there are other ways of making the cuts1 but this rule must always be observed if we are to get the full number of pieces4 ,the general formula is that with n cuts we can always produce 77n7n#a77_/b7a 4 ,one of the problems proposed by the late ,sam ,loyd was to produce the maximum number of pieces by n straight cuts through a solid cheese4 ,of course1 again1 the pieces cut off may not be moved or piled4 ,here we have to deal with the intersection of planes 7instead of lines71 and the general formula is that with n cuts we may produce 77n - #a7n7n #a77_/f n #a pieces4 ,it is extremely difficult to see the direction and effects of the successive cuts for more than a few of the lowest values of n4 #afe4--,,the ,,seven ,,pigs4 ,the illustration shows the direction for placing the three fences so as to enclose every pig in a separate sty4 ,the greatest number of spaces that can be enclosed with three straight lines in a square is seven1 as shown in the last puzzle4 ,bearing this fact in mind1 the puzzle must be solved by trial4 ,illustration3 ,,the ,,seven ,,pigs4 #aff4--,,the ,,landowner',s ,,fences4 ,four fences only are necessary1 as follows3-- ,illustration #afg4--,,the ,,wizard',s ,,cats4 ,the illustration requires no explanation4 ,it shows clearly how the three circles may be drawn so that every cat has a separate enclosure1 and cannot approach another cat without crossing a line4 ,illustration3 ,,the ,,wizards' ,,cats4 #afh4--,,the ,,christmas ,,pudding4 ,the illustration shows how the pudding may be cut into two parts of exactly the same size and shape4 ,the lines must necessarily pass through the points ,a1 ,b1 ,c1 ,d1 and ,e4 ,but1 subject to this condition1 they may be varied in an infinite number of ways4 ,for example1 at a point midway between ,a and the edge1 the line may be completed in an unlimited number of ways 7straight or crooked71 provided it be exactly reflected from ,e to the opposite edge4 ,and similar variations may be introduced at other places4 ,illustration #afi4--;,a ,,tangram ,,paradox4 ,the diagrams will show how the figures are constructed--each with the seven ,tangrams4 ,it will be noticed that in both cases the head1 hat1 and arm are precisely alike1 and the width at the base of the body the same4 ,but this body contains four pieces in the first case1 and in the second design only three4 ,the first is larger than the second by exactly that narrow strip indicated by the dotted line between ,a and ,b4 ,this strip is therefore exactly equal in area to the piece forming the foot in the other design1 though when thus distributed along the side of the body the increased dimension is not easily apparent to the eye4 ,illustration #agj4--,,the ,,cushion ,,covers4 ,illustration ,the two pieces of brocade marked ,a will fit together and form one perfect square cushion top1 and the two pieces marked ,b will form the other4 #aga4--,,the ,,banner ,,puzzle4 ,the illustration explains itself4 ,divide the bunting into #be squares 7because this number is the sum of two other squares--#af and #i71 and then cut along the thick lines4 ,the two pieces marked ,a form one square1 and the two pieces marked ,b form the other4 ,illustration #agb4--,,mrs4 ,,smiley',s ,,christmas ,,present4 ,illustration ,illustration ,the first step is to find six different square numbers that sum to #aif4 ,for example1 #a #d #be #cf #di #ha #aif2 #a #d #i #be #cf #aba #aif2 #a #i #af #be #fd #ha #aif4 ,the rest calls for individual judgment and ingenuity1 and no definite rules can be given for procedure4 ,the annexed diagrams will show solutions for the first two cases stated4 ,of course the three pieces marked ,a and those marked ,b will fit together and form a square in each case4 ,the assembling of the parts may be slightly varied1 and the reader may be interested in finding a solution for the third set of squares ,i have given4 #agc4--,,mrs4 ,,perkins',s ,,quilt4 ,the following diagram shows how the quilt should be constructed4 ,illustration ,there is1 ,i believe1 practically only one solution to this puzzle4 ,the fewest separate squares must be eleven4 ,the portions must be of the sizes given1 the three largest pieces must be arranged as shown1 and the remaining group of eight squares may be reflected1 but cannot be differently arranged4 #agd4--,,the ,,squares ,,of ,,brocade4 ,illustration3 ,diagram #a ,so far as ,i have been able to discover1 there is only one possible solution to fulfil the conditions4 ,the pieces fit together as in ,diagram #a1 ,diagrams #b and #c showing how the two original squares are to be cut4 ,it will be seen that the pieces ,a and ,c have each twenty chequers1 and are therefore of equal area4 ,diagram #d 7built up with the dissected square ,no4 #e7 solves the puzzle1 except for the small condition contained in the words1 ,i cut the two squares in the manner desired4 ,in this case the smaller square is preserved intact4 ,still ,i give it as an illustration of a feature of the puzzle4 ,it is impossible in a problem of this kind to give a quarter-turn to any of the pieces if the pattern is to properly match1 but 7as in the case of ,f1 in ,diagram #d7 we may give a symmetrical piece a half-turn--that is1 turn it upside down4 ,whether or not a piece may be given a quarter-turn1 a half-turn1 or no turn at all in these chequered problems1 depends on the character of the design1 on the material employed1 and also on the form of the piece itself4 ,illustration3 ,diagram #b ,illustration3 ,diagram #c ,illustration3 ,diagram #d ,illustration3 ,diagram #e #age4--;,,another ,,patchwork ,,puzzle4 ,the lady need only unpick the stitches along the dark lines in the larger portion of patchwork1 when the four pieces will fit together and form a square1 as shown in our illustration4 ,illustration #agf4--,,linoleum ,,cutting4 ,there is only one solution that will enable us to retain the larger of the two pieces with as little as possible cut from it4 ,fig4 #a in the following diagram shows how the smaller piece is to be cut1 and ,fig4 #b how we should dissect the larger piece1 while in ,fig4 #c we have the new square #aj #aj formed by the four pieces with all the chequers properly matched4 ,it will be seen that the piece ,d contains fifty-two chequers1 and this is the largest piece that it is possible to preserve under the conditions4 ,illustration #agg4--;,,another ,,linoleum ,,puzzle4 ,cut along the thick lines1 and the four pieces will fit together and form a perfect square in the manner shown in the smaller diagram4 ,illustration3 ,,another ,,linoleum ,,puzzle4 #agh4--,,the ,,cardboard ,,box4 ,the areas of the top and side multiplied together and divided by the area of the end give the square of the length4 ,similarly1 the product of top and end divided by side gives the square of the breadth2 and the product of side and end divided by the top gives the square of the depth4 ,but we only need one of these operations4 ,let us take the first4 ,thus1 #abj #if divided by #hj equals #add1 the square of #ab4 ,therefore the length is #ab inches1 from which we can1 of course1 at once get the breadth and depth--#aj in4 and #h in4 respectively4 #agi4--,,stealing ,,the ,,bell-,,ropes4 ,whenever we have one side 7a7 of a right-angled triangle1 and know the difference between the second side and the hypotenuse 7which difference we will call b71 then the length of the hypotenuse will be a b --- -4 #b;b #b ,in the case of our puzzle this will be #dh #dh ------- #a in4 #cb ft4 #a in41 #f which is the length of the rope4 #ahj-- ,,the ,,four ,,sons4 ,illustration ,the diagram shows the most equitable division of the land possible1 so that each son shall receive land of exactly the same area and exactly similar in shape1 and so that each shall have access to the well in the centre without trespass on another's land4 ,the conditions do not require that each son's land shall be in one piece1 but it is necessary that the two portions assigned to an individual should be kept apart1 or two adjoining portions might be held to be one piece1 in which case the condition as to shape would have to be broken4 ,at present there is only one shape for each piece of land--half a square divided diagonally4 ,and ,a1 ,b1 ,c1 and ,d can each reach their land from the outside1 and have each equal access to the well in the centre4 #aha4--,,the ,,three ,,railway ,,stations4 ,the three stations form a triangle1 with sides #ac1 #ad1 and #ae miles4 ,make the #ad side the base2 then the height of the triangle is #ab and the area #hd4 ,multiply the three sides together and divide by four times the area4 ,the result is eight miles and one-eighth1 the distance required4 #ahb4--,,the ,,garden ,,puzzle4 ,half the sum of the four sides is #add4 ,from this deduct in turn the four sides1 and we get #fd1 #ii1 #dd1 and #ha4 ,multiply these together1 and we have as the result the square of #d1geb4 ,therefore the garden contained #d1geb square yards4 ,of course the tree being equidistant from the four corners shows that the garden is a quadrilateral that may be inscribed in a circle4 #ahc4--;,,drawing ,a ,,spiral4 ,make a fold in the paper1 as shown by the dotted line in the illustration4 ,then1 taking any two points1 as ,a and ,b1 describe semicircles on the line alternately from the centres ,b and ,a1 being careful to make the ends join1 and the thing is done4 ,of course this is not a true spiral1 but the puzzle was to produce the particular spiral that was shown1 and that was drawn in this simple manner4 ,illustration #ahd4--;,,how ,,to ,,draw ,,an ,,oval4 ,if you place your sheet of paper round the surface of a cylindrical bottle or canister1 the oval can be drawn with one sweep of the compasses4 #ahe4--,,st4 ,,george',s ,,banner4 ,as the flag measures #d ft4 by #c ft41 the length of the diagonal 7from corner to corner7 is #e ft4 ,all you need do is to deduct half the length of this diagonal 7#b ft47 from a quarter of the distance all round the edge of the flag 7#c ft47--a quarter of #ad ft4 ,the difference 7#a ft47 is the required width of the arm of the red cross4 ,the area of the cross will then be the same as that of the white ground4 #ahf4--,,the ,,clothes ,,line ,,puzzle4 ,multiply together1 and also add together1 the heights of the two poles and divide one result by the other4 ,that is1 if the two heights are a and b respectively1 then ab_/7a b7 will give the height of the intersection4 ,in the particular case of our puzzle1 the intersection was therefore #b ft4 #aa in4 from the ground4 ,the distance that the poles are apart does not affect the answer4 ,the reader who may have imagined that this was an accidental omission will perhaps be interested in discovering the reason why the distance between the poles may be ignored4 #ahg4--,,the ,,milkmaid ,,puzzle4 ,illustration3 ,a ,b ,,river ------------------ _/ _/ _/ _/ ,,door ,,stool ,draw a straight line1 as shown in the diagram1 from the milking-stool perpendicular to the near bank of the river1 and continue it to the point ,a1 which is the same distance from that bank as the stool4 ,if you now draw the straight line from ,a to the door of the dairy1 it will cut the river at ,b4 ,then the shortest route will be from the stool to ,b and thence to the door4 ,obviously the shortest distance from ,a to the door is the straight line1 and as the distance from the stool to any point of the river is the same as from ,a to that point1 the correctness of the solution will probably appeal to every reader without any acquaintance with geometry4 #ahh4--,,the ,,ball ,,problem4 ,if a round ball is placed on the level ground1 six similar balls may be placed round it 7all on the ground71 so that they shall all touch the central ball4 ,as for the second question1 the ratio of the diameter of a circle to its circumference we call pi2 and though we cannot express this ratio in exact numbers1 we can get sufficiently near to it for all practical purposes4 ,however1 in this case it is not necessary to know the value of pi at all4 ,because1 to find the area of the surface of a sphere we multiply the square of the diameter by pi2 to find the volume of a sphere we multiply the cube of the diameter by one-sixth of pi4 ,therefore we may ignore pi1 and have merely to seek a number whose square shall equal one-sixth of its cube4 ,this number is obviously #f4 ,therefore the ball was #f ft4 in diameter1 for the area of its surface will be #cf times pi in square feet1 and its volume also #cf times pi in cubic feet4 #ahi4--,,the ,,yorkshire ,,estates4 ,the triangular piece of land that was not for sale contains exactly eleven acres4 ,of course it is not difficult to find the answer if we follow the eccentric and tricky tracks of intricate trigonometry2 or ,i might say that the application of a well-known formula reduces the problem to finding one-quarter of the square root of 7#d #cgj #aaf7 -7#cgj #aaf - #gd7--that is a quarter of the square root of #aicf1 which is one-quarter of #dd1 or #aa acres4 ,but all that the reader really requires to know is the ,pythagorean law on which many puzzles have been built1 that in any right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides4 ,i shall dispense with all surds and similar absurdities1 notwithstanding the fact that the sides of our triangle are clearly incommensurate1 since we cannot exactly extract the square roots of the three square areas4 ,illustration3 ,a 4 4 #e 4 #g 4 ,e --------- ,c 4 4 4 4 4 #d #d 4 4 #g 44 ,d--------------------------- ,b ,f ,in the above diagram ,,abc represents our triangle4 ,,adb is a right-angled triangle1 ,,ad measuring #i and ,,bd measuring #ag1 because the square of #i added to the square of #ag equals #cgj1 the known area of the square on ,,ab4 ,also ,,aec is a right-angled triangle1 and the square of #e added to the square of #g equals #gd1 the square estate on ,a ,c4 ,similarly1 ,,cfb is a right-angled triangle1 for the square of #d added to the square of #aj equals #aaf1 the square estate on ,,bc4 ,now1 although the sides of our triangular estate are incommensurate1 we have in this diagram all the exact figures that we need to discover the area with precision4 ,the area of our triangle ,,adb is clearly half of #i #ag1 or #gf acres4 ,the area of ,,aec is half of #e #g1 or #ag acres2 the area of ,,cfb is half of #d #aj1 or #bj acres2 and the area of the oblong ,,edfc is obviously #d #g1 or #bh acres4 ,now1 if we add together #ag1 #bj1 and #bh #fe1 and deduct this sum from the area of the large triangle ,,adb 7which we have found to be #gf acres71 what remains must clearly be the area of ,,abc4 ,that is to say1 the area we want must be #gf - #fe #aa acres exactly4 #aij4--;,,farmer ,,wurzel',s ,,estate4 ,the area of the complete estate is exactly one hundred acres4 ,to find this answer ,i use the following little formula1 _/#d;ab - 7a b - c7 -------------------- #d where a1 b1 c represent the three square areas1 in any order4 ,the expression gives the area of the triangle ,a4 ,this will be found to be #i acres4 ,it can be easily proved that ,a1 ,b1 ,c1 and ,d are all equal in area2 so the answer is #bf #bj #ah #i #i #i #i #ajj acres4 ,illustration ,here is the proof4 ,if every little dotted square in the diagram represents an acre1 this must be a correct plan of the estate1 for the squares of #e and #a together equal #bf2 the squares of #d and #b equal #bj2 and the squares of #c and #c added together equal #ah4 ,now we see at once that the area of the triangle ,e is #b1 ,f is #d1 and ,g is #d4 ,these added together make #aa acres1 which we deduct from the area of the rectangle1 #bj acres1 and we find that the field ,a contains exactly #i acres4 ,if you want to prove that ,b1 ,c1 and ,d are equal in size to ,a1 divide them in two by a line from the middle of the longest side to the opposite angle1 and you will find that the two pieces in every case1 if cut out1 will exactly fit together and form ,a4 ,or we can get our proof in a still easier way4 ,the complete area of the squared diagram is #ab #ab #add acres1 and the portions #a1 #b1 #c1 #d1 not included in the estate1 have the respective areas of #ab1 #ag1 #i1 and #d4 ,these added together make #dd1 which1 deducted from #add1 leaves #ajj as the required area of the complete estate4 #aia4--,,the ,,crescent ,,puzzle4 ,referring to the original diagram1 let ,,ac be x1 let ,,cd be x - #i1 and let ,,ec be x - #e4 ,then x - #e is a mean proportional between x - #i and x1 from which we find that x equals #be4 ,therefore the diameters are #ej in4 and #da in4 respectively4 #aib4--,,the ,,puzzle ,,wall4 ,illustration ,the answer given in all the old books is that shown in ,fig4 #a1 where the curved wall shuts out the cottages from access to the lake4 ,but in seeking the direction for the shortest possible wall most readers to-day1 remembering that the shortest distance between two points is a straight line1 will adopt the method shown in ,fig4 #b4 ,this is certainly an improvement1 yet the correct answer is really that indicated in ,fig4 #c4 ,a measurement of the lines will show that there is a considerable saving of length in this wall4 #aic4--,,the ,,sheep-,,fold4 ,this is the answer that is always given and accepted as correct3 ,two more hurdles would be necessary1 for the pen was twenty-four by one 7as in ,fig4 ,a on next page71 and by moving one of the sides and placing an extra hurdle at each end 7as in ,fig4 ,b7 the area would be doubled4 ,the diagrams are not to scale4 ,now there is no condition in the puzzle that requires the sheep-fold to be of any particular form4 ,but even if we accept the point that the pen was twenty-four by one1 the answer utterly fails1 for two extra hurdles are certainly not at all necessary4 ,for example1 ,i arrange the fifty hurdles as in ,fig4 ,c1 and as the area is increased from twenty-four square hurdles to #aef1 there is now accommodation for #fej sheep4 ,if it be held that the area must be exactly double that of the original pen1 then ,i construct it 7as in ,fig4 ,d7 with twenty-eight hurdles only1 and have twenty-two in hand for other purposes on the farm4 ,even if it were insisted that all the original hurdles must be used1 then ,i should construct it as in ,fig4 ,e1 where ,i can get the area as exact as any farmer could possibly require1 even if we have to allow for the fact that the sheep might not be able to graze at the extreme ends4 ,thus we see that1 from any point of view1 the accepted answer to this ancient little puzzle breaks down4 ,and yet attention has never before been drawn to the absurdity4 ,illustration ,a #bd -------------------------------- #bd #a -------------------------------- ,b -------------------------------- #dh #b -------------------------------- #bd ,c -------------------- ,d ---------- #ab #dh #f #aef ---------- #h -------------------- #ac #ab 4 ,e #ac 4 ' ' 4 4 ' ' 4 ' 4 4 ' #ab ' 4 ' #ac #aid4--,,the ,,garden ,,walls4 ,the puzzle was to divide the circular field into four equal parts by three walls1 each wall being of exactly the same length4 ,there are two essential difficulties in this problem4 ,these are3 7#a7 the thickness of the walls1 and 7#b7 the condition that these walls are three in number4 ,as to the first point1 since we are told that the walls are brick walls1 we clearly cannot ignore their thickness1 while we have to find a solution that will equally work1 whether the walls be of a thickness of one1 two1 three1 or more bricks4 ,illustration ,the second point requires a little more consideration4 ,how are we to distinguish between a wall and walls8 ,a straight wall without any bend in it1 no matter how long1 cannot ever become walls1 if it is neither broken nor intersected in any way4 ,also our circular field is clearly enclosed by one wall4 ,but if it had happened to be a square or a triangular enclosure1 would there be respectively four and three walls or only one enclosing wall in each case8 ,it is true that we speak of the four walls of a square building or garden1 but this is only a conventional way of saying the four sides4 ,if you were speaking of the actual brickwork1 you would say1 ,i am going to enclose this square garden with a wall4 ,angles clearly do not affect the question1 for we may have a zigzag wall just as well as a straight one1 and the ,great ,wall of ,china is a good example of a wall with plenty of angles4 ,now1 if you look at ,diagrams #a1 #b1 and #c1 you may be puzzled to declare whether there are in each case two or four new walls2 but you cannot call them three1 as required in our puzzle4 ,the intersection either affects the question or it does not affect it4 ,if you tie two pieces of string firmly together1 or splice them in a nautical manner1 they become one piece of string4 ,if you simply let them lie across one another or overlap1 they remain two pieces of string4 ,it is all a question of joining and welding4 ,it may similarly be held that if two walls be built into one another--,i might almost say1 if they be made homogeneous--they become one wall1 in which case ,diagrams #a1 #b1 and #c might each be said to show one wall or two1 if it be indicated that the four ends only touch1 and are not really built into1 the outer circular wall4 ,the objection to ,diagram #d is that although it shows the three required walls 7assuming the ends are not built into the outer circular wall71 yet it is only absolutely correct when we assume the walls to have no thickness4 ,a brick has thickness1 and therefore the fact throws the whole method out and renders it only approximately correct4 ,diagram #e shows1 perhaps1 the only correct and perfectly satisfactory solution4 ,it will be noticed that1 in addition to the circular wall1 there are three new walls1 which touch 7and so enclose7 but are not built into one another4 ,this solution may be adapted to any desired thickness of wall1 and its correctness as to area and length of wall space is so obvious that it is unnecessary to explain it4 ,i will1 however1 just say that the semicircular piece of ground that each tenant gives to his neighbour is exactly equal to the semicircular piece that his neighbour gives to him1 while any section of wall space found in one garden is precisely repeated in all the others4 ,of course there is an infinite number of ways in which this solution may be correctly varied4 #aie4--,,lady ,,belinda',s ,,garden4 ,all that ,lady ,belinda need do was this3 ,she should measure from ,a to ,b1 fold her tape in four and mark off the point ,e1 which is thus one quarter of the side4 ,then1 in the same way1 mark off the point ,f1 one-fourth of the side ,,ad ,now1 if she makes ,,eg equal to ,,af1 and ,,gh equal to ,,ef1 then ,,ah is the required width for the path in order that the bed shall be exactly half the area of the garden4 ,an exact numerical measurement can only be obtained when the sum of the squares of the two sides is a square number4 ,thus1 if the garden measured #ab poles by #e poles 7where the squares of #ab and #e1 #add and #be1 sum to #afi1 the square of #ac71 then #ab added to #e1 less #ac1 would equal four1 and a quarter of this1 #a pole1 would be the width of the path4 #aif4--,,the ,,tethered ,,goat4 ,illustration ,this problem is quite simple if properly attacked4 ,let us suppose the triangle ,,abc to represent our half-acre field1 and the shaded portion to be the quarter-acre over which the goat will graze when tethered to the corner ,c4 ,now1 as six equal equilateral triangles placed together will form a regular hexagon1 as shown1 it is evident that the shaded pasture is just one-sixth of the complete area of a circle4 ,therefore all we require is the radius 7,,cd7 of a circle containing six quarter-acres or #a acres1 which is equal to #i1djh1ifj square inches4 ,as we only want our answer to the nearest inch1 it is sufficiently exact for our purpose if we assume that as #a is to #c4adaf1 so is the diameter of a circle to its circumference4 ,if1 therefore1 we divide the last number ,i gave by #c4adaf1 and extract the square root1 we find that #a1gca inches1 or #dh yards #c inches1 is the required length of the tether to the nearest inch4 #aig4--,,the ,,compasses ,,puzzle4 ,let ,,ab in the following diagram be the given straight line4 ,with the centres ,a and ,b and radius ,,ab describe the two circles4 ,mark off ,,de and ,,ef equal to ,,ad4 ,with the centres ,a and ,f and radius ,,df describe arcs intersecting at ,g4 ,with the centres ,a and ,b and distance ,,bg describe arcs ,,ghk and ,n4 ,make ,,hk equal to ,,ab and ,,hl equal to ,,hb4 ,then with centres ,k and ,l and radius ,,ab describe arcs intersecting at ,i4 ,make ,,bm equal to ,,bi4 ,finally1 with the centre ,m and radius ,,mb cut the line in ,c1 and the point ,c is the required middle of the line ,,ab4 ,for greater exactitude you can mark off ,r from ,a 7as you did ,m from ,b71 and from ,r describe another arc at ,c4 ,this also solves the problem1 to find a point midway between two given points without the straight line4 ,illustration ,i will put the young geometer in the way of a rigid proof4 ,first prove that twice the square of the line ,,ab equals the square of the distance ,,bg1 from which it follows that ,,habn are the four corners of a square4 ,to prove that ,i is the centre of this square1 draw a line from ,h to ,p through ,,qib and continue the arc ,,hk to ,p4 ,then1 conceiving the necessary lines to be drawn1 the angle ,,hkp1 being in a semicircle1 is a right angle4 ,let fall the perpendicular ,,kq1 and by similar triangles1 and from the fact that ,,hki is an isosceles triangle by the construction1 it can be proved that ,,hi is half of ,,hb4 ,we can similarly prove that ,c is the centre of the square of which ,,aib are three corners4 ,i am aware that this is not the simplest possible solution4 #aih4--,,the ,,eight ,,sticks4 ,the first diagram is the answer that nearly every one will give to this puzzle1 and at first sight it seems quite satisfactory4 ,but consider the conditions4 ,we have to lay every one of the sticks on the table4 ,now1 if a ladder be placed against a wall with only one end on the ground1 it can hardly be said that it is laid on the ground4 ,and if we place the sticks in the above manner1 it is only possible to make one end of two of them touch the table3 to say that every one lies on the table would not be correct4 ,to obtain a solution it is only necessary to have our sticks of proper dimensions4 ,say the long sticks are each #b ft4 in length and the short ones #a ft4 ,then the sticks must be #c in4 thick1 when the three equal squares may be enclosed1 as shown in the second diagram4 ,if ,i had said matches instead of sticks1 the puzzle would be impossible1 because an ordinary match is about twenty-one times as long as it is broad1 and the enclosed rectangles would not be squares4 ,illustration #aii4--,,papa',s ,,puzzle4 ,i have found that a large number of people imagine that the following is a correct solution of the problem4 ,using the letters in the diagram below1 they argue that if you make the distance ,,ba one-third of ,,bc1 and therefore the area of the rectangle ,,abe equal to that of the triangular remainder1 the card must hang with the long side horizontal4 ,readers will remember the jest of ,charles ,,ii41 who induced the ,royal ,society to meet and discuss the reason why the water in a vessel will not rise if you put a live fish in it2 but in the middle of the proceedings one of the least distinguished among them quietly slipped out and made the experiment1 when he found that the water did rise6 ,if my correspondents had similarly made the experiment with a piece of cardboard1 they would have found at once their error4 ,area is one thing1 but gravitation is quite another4 ,the fact of that triangle sticking its leg out to ,d has to be compensated for by additional area in the rectangle4 ,as a matter of fact1 the ratio of ,,ba to ,,ac is as #a is to the square root of #c1 which latter cannot be given in an exact numerical measure1 but is approximately #a4gcb4 ,now let us look at the correct general solution4 ,there are many ways of arriving at the desired result1 but the one ,i give is1 ,i think1 the simplest for beginners4 ,illustration ,fix your card on a piece of paper and draw the equilateral triangle ,,bcf1 ,,bf and ,,cf being equal to ,,bc4 ,also mark off the point ,g so that ,,dg shall equal ,,dc4 ,draw the line ,,cg and produce it until it cuts the line ,,bf in ,h4 ,if we now make ,,ha parallel to ,,be1 then ,a is the point from which our cut must be made to the corner ,d1 as indicated by the dotted line4 ,a curious point in connection with this problem is the fact that the position of the point ,a is independent of the side ,,cd4 ,the reason for this is more obvious in the solution ,i have given than in any other method that ,i have seen1 and 7although the problem may be solved with all the working on the cardboard7 that is partly why ,i have preferred it4 ,it will be seen at once that however much you may reduce the width of the card by bringing ,e nearer to ,b and ,d nearer to ,c1 the line ,,cg1 being the diagonal of a square1 will always lie in the same direction1 and will cut ,,bf in ,h4 ,finally1 if you wish to get an approximate measure for the distance ,,ba1 all you have to do is to multiply the length of the card by the decimal 4#cff4 ,thus1 if the card were #g inches long1 we get #g 4#cff #b4efb1 or a little more than #b inches1 for the distance from ,b to ,a4 ,but the real joke of the puzzle is this3 ,we have seen that the position of the point ,a is independent of the width of the card1 and depends entirely on the length4 ,now1 in the illustration it will be found that both cards have the same length2 consequently all the little maid had to do was to lay the clipped card on top of the other one and mark off the point ,a at precisely the same distance from the top left-hand corner6 ,so1 after all1 ,pappus' puzzle1 as he presented it to his little maid1 was quite an infantile problem1 when he was able to show her how to perform the feat without first introducing her to the elements of statics and geometry4 #bjj4--;,a ,,kite-,,flying ,,puzzle4 ,solvers of this little puzzle1 ,i have generally found1 may be roughly divided into two classes3 those who get within a mile of the correct answer by means of more or less complex calculations1 involving pi1 and those whose arithmetical kites fly hundreds and thousands of miles away from the truth4 ,the comparatively easy method that ,i shall show does not involve any consideration of the ratio that the diameter of a circle bears to its circumference4 ,i call it the hat-box method4 ,illustration ,supposing we place our ball of wire1 ,a1 in a cylindrical hat-box1 ,b1 that exactly fits it1 so that it touches the side all round and exactly touches the top and bottom1 as shown in the illustration4 ,then1 by an invariable law that should be known by everybody1 that box contains exactly half as much again as the ball4 ,therefore1 as the ball is #bd in4 in diameter1 a hat-box of the same circumference but two-thirds of the height 7that is1 #af in4 high7 will have exactly the same contents as the ball4 ,now let us consider that this reduced hat-box is a cylinder of metal made up of an immense number of little wire cylinders close together like the hairs in a painter's brush4 ,by the conditions of the puzzle we are allowed to consider that there are no spaces between the wires4 ,how many of these cylinders one one-hundredth of an inch thick are equal to the large cylinder1 which is #bd in4 thick8 ,circles are to one another as the squares of their diameters4 ,the square of #a_/ajj is #a_/ajjjjj1 and the square of #bd is #egf2 therefore the large cylinder contains #e1gfj1jjj of the little wire cylinders4 ,but we have seen that each of these wires is #af in4 long2 hence #af #e1gfj1jjj #ib1afj1jjj inches as the complete length of the wire4 ,reduce this to miles1 and we get #a1ded miles #b1hhj ft4 as the length of the wire attached to the professor's kite4 ,whether a kite would fly at such a height1 or support such a weight1 are questions that do not enter into the problem4 #bja4--;,,how ,,to ,,make ,,cisterns4 ,here is a general formula for solving this problem4 ,call the two sides of the rectangle a and b4 ,then a b - 7a b - ab7 --------------------------- #f equals the side of the little square pieces to cut away4 ,the measurements given were #h ft4 by #c ft41 and the above rule gives #h in4 as the side of the square pieces that have to be cut away4 ,of course it will not always come out exact1 as in this case 7on account of that square root71 but you can get as near as you like with decimals4 #bjb4--,,the ,,cone ,,puzzle4 ,the simple rule is that the cone must be cut at one-third of its altitude4 #bjc4--;,,concerning ,,wheels4 ,if you mark a point ,a on the circumference of a wheel that runs on the surface of a level road1 like an ordinary cart-wheel1 the curve described by that point will be a common cycloid1 as in ,fig4 #a4 ,but if you mark a point ,b on the circumference of the flange of a locomotive-wheel1 the curve will be a curtate cycloid1 as in ,fig4 #b1 terminating in nodes4 ,now1 if we consider one of these nodes or loops1 we shall see that at any given moment certain points at the bottom of the loop must be moving in the opposite direction to the train4 ,as there is an infinite number of such points on the flange's circumference1 there must be an infinite number of these loops being described while the train is in motion4 ,in fact1 at any given moment certain points on the flanges are always moving in a direction opposite to that in which the train is going4 ,illustration3 #a ,illustration3 #b ,in the case of the two wheels1 the wheel that runs round the stationary one makes two revolutions round its own centre4 ,as both wheels are of the same size1 it is obvious that if at the start we mark a point on the circumference of the upper wheel1 at the very top1 this point will be in contact with the lower wheel at its lowest part when half the journey has been made4 ,therefore this point is again at the top of the moving wheel1 and one revolution has been made4 ,consequently there are two such revolutions in the complete journey4 #bjd4--;,a ,,new ,,match ,,puzzle4 #a4 ,the easiest way is to arrange the eighteen matches as in ,diagrams #a and #b1 making the length of the perpendicular ,,ab equal to a match and a half4 ,then1 if the matches are an inch in length1 ,fig4 #a contains two square inches and ,fig4 #b contains six square inches--#d #a4 ,the second case 7#b7 is a little more difficult to solve4 ,the solution is given in ,figs4 #c and #d4 ,for the purpose of construction1 place matches temporarily on the dotted lines4 ,then it will be seen that as #c contains five equal equilateral triangles and #d contains fifteen similar triangles1 one figure is three times as large as the other1 and exactly eighteen matches are used4 ,illustration3 ,figures #a1 #b1 #c1 #d4 #bje4--,,the ,,six ,,sheep-,,pens4 ,illustration ,place the twelve matches in the manner shown in the illustration1 and you will have six pens of equal size4 #bjf4--,,the ,,king ,,and ,,the ,,castles4 ,there are various ways of building the ten castles so that they shall form five rows with four castles in every row1 but the arrangement in the next column is the only one that also provides that two castles 7the greatest number possible7 shall not be approachable from the outside4 ,it will be seen that you must cross the walls to reach these two4 ,illustration3 ,the ,king and the ,castles #bjg4--;,,cherries ,,and ,,plums4 ,there are several ways in which this problem might be solved were it not for the condition that as few cherries and plums as possible shall be planted on the north and east sides of the orchard4 ,the best possible arrangement is that shown in the diagram1 where the cherries1 plums1 and apples are indicated respectively by the letters ,c1 ,p1 and ,a4 ,the dotted lines connect the cherries1 and the other lines the plums4 ,it will be seen that the ten cherry trees and the ten plum trees are so planted that each fruit forms five lines with four trees of its kind in line4 ,this is the only arrangement that allows of so few as two cherries or plums being planted on the north and east outside rows4 ,illustration #bjh4--;,a ,,plantation ,,puzzle4 ,the illustration shows the ten trees that must be left to form five rows with four trees in every row4 ,the dots represent the positions of the trees that have been cut down4 ,illustration #bji4--,,the ,,twenty-,,one ,,trees4 ,i give two pleasing arrangements of the trees4 ,in each case there are twelve straight rows with five trees in every row4 ,illustration3 ,figure #a1 ,figure #b4 #baj4--,,the ,,ten ,,coins4 ,the answer is that there are just #b1djj different ways4 ,any three coins may be taken from one side to combine with one coin taken from the other side4 ,i give four examples on this and the next page4 ,we may thus select three from the top in ten ways and one from the bottom in five ways1 making fifty4 ,but we may also select three from the bottom and one from the top in fifty ways4 ,we may thus select the four coins in one hundred ways1 and the four removed may be arranged by permutation in twenty-four ways4 ,thus there are #bd #ajj #b1djj different solutions4 ,illustration ,as all the points and lines puzzles that ,i have given so far1 excepting the last1 are variations of the case of ten points arranged to form five lines of four1 it will be well to consider this particular case generally4 ,there are six fundamental solutions1 and no more1 as shown in the six diagrams4 ,these1 for the sake of convenience1 ,i named some years ago the ,star1 the ,dart1 the ,compasses1 the ,funnel1 the ,scissors1 and the ,nail4 7,see next page47 ,readers will understand that any one of these forms may be distorted in an infinite number of different ways without destroying its real character4 ,in ,the ,king and the ,castles we have the ,star1 and its solution gives the ,compasses4 ,in the ,cherries and ,plums solution we find that the ,cherries represent the ,funnel and the ,plums the ,dart4 ,the solution of the ,plantation ,puzzle is an example of the ,dart distorted4 ,any solution to the ,ten ,coins will represent the ,scissors4 ,thus examples of all have been given except the ,nail4 ,on a reduced chessboard1 #g by #g1 we may place the ten pawns in just three different ways1 but they must all represent the ,dart4 ,the ,plantation shows one way1 the ,plums show a second way1 and the reader may like to find the third way for himself4 ,on an ordinary chessboard1 #h by #h1 we can also get in a beautiful example of the ,funnel--symmetrical in relation to the diagonal of the board4 ,the smallest board that will take a ,star is one #i by #g4 ,the ,nail requires a board #aa by #g1 the ,scissors ,illustration #aa by #i1 and the ,compasses #ag by #ab4 ,at least these are the best results recorded in my note-book4 ,they may be beaten1 but ,i do not think so4 ,if you divide a chessboard into two parts by a diagonal zigzag line1 so that the larger part contains #cf squares and the smaller part #bh squares1 you can place three separate schemes on the larger part and one on the smaller part 7all ,darts7 without their conflicting--that is1 they occupy forty different squares4 ,they can be placed in other ways without a division of the board4 ,the smallest square board that will contain six different schemes 7not fundamentally different71 without any line of one scheme crossing the line of another1 is #ad by #ad2 and the smallest board that will contain one scheme entirely enclosed within the lines of a second scheme1 without any of the lines of the one1 when drawn from point to point1 crossing a line of the other1 is #ad by #ab4 ,illustration3 ,,star ,,dart ,,compasses ,,funnel ,,scissors ,,nail #baa4--,,the ,,twelve ,,mince-,,pies4 ,if you ignore the four black pies in our illustration1 the remaining twelve are in their original positions4 ,now remove the four detached pies to the places occupied by the black ones1 and you will have your seven straight rows of four1 as shown by the dotted lines4 ,illustration3 ,the ,twelve ,mince ,pies4 #bab4--,,the ,,burmese ,,plantation4 ,the arrangement on the next page is the most symmetrical answer that can probably be found for twenty-one rows1 which is1 ,i believe1 the greatest number of rows possible4 ,there are several ways of doing it4 #bac4--,,turks ,,and ,,russians4 ,the main point is to discover the smallest possible number of ,russians that there could have been4 ,as the enemy opened fire from all directions1 it is clearly necessary to find what is the smallest number of heads that could form sixteen lines with three heads in every line4 ,note that ,i say sixteen1 and not thirty-two1 because every line taken by a bullet may be also taken by another bullet fired in exactly the opposite direction4 ,now1 as few as eleven points1 or heads1 may be arranged to form the required sixteen lines of three1 but the discovery of this arrangement is a hard nut4 ,the diagram at the foot of this page will show exactly how the thing is to be done4 ,illustration ,if1 therefore1 eleven ,russians were in the positions shown by the stars1 and the thirty-two ,turks in the positions indicated by the black dots1 it will be seen1 by the lines shown1 that each ,turk may fire exactly over the heads of three ,russians4 ,but as each bullet kills a man1 it is essential that every ,turk shall shoot one of his comrades and be shot by him in turn2 otherwise we should have to provide extra ,russians to be shot1 which would be destructive of the correct solution of our problem4 ,as the firing was simultaneous1 this point presents no difficulties4 ,the answer we thus see is that there were at least eleven ,russians amongst whom there was no casualty1 and that all the thirty-two ,turks were shot by one another4 ,it was not stated whether the ,russians fired any shots1 but it will be evident that even if they did their firing could not have been effective3 for if one of their bullets killed a ,turk1 then we have immediately to provide another man for one of the ,turkish bullets to kill2 and as the ,turks were known to be thirty-two in number1 this would necessitate our introducing another ,russian soldier and1 of course1 destroying the solution4 ,i repeat that the difficulty of the puzzle consists in finding how to arrange eleven points so that they shall form sixteen lines of three4 ,i am told that the possibility of doing this was first discovered by the ,rev4 ,mr4 ,wilkinson some twenty years ago4 #bad4--,,the ,,six ,,frogs4 ,move the frogs in the following order3 #b1 #d1 #f1 #e1 #c1 #a 7repeat these moves in the same order twice more71 #b1 #d1 #f4 ,this is a solution in twenty-one moves--the fewest possible4 ,if n1 the number of frogs1 be even1 we require 7n n7_/#b moves1 of which 7n - n7_/#b will be leaps and n simple moves4 ,if n be odd1 we shall need 77n #cn7_/#b7 - #d moves1 of which 7n - n7_/#b will be leaps and #bn - #d simple moves4 ,in the even cases write1 for the moves1 all the even numbers in ascending order and the odd numbers in descending order4 ,this series must be repeated n times and followed by the even numbers in ascending order once only4 ,thus the solution for #ad frogs will be 7#b1 #d1 #f1 #h1 #aj1 #ab1 #ad1 #ac1 #aa1 #i1 #g1 #e1 #c1 #a7 repeated #g times and followed by #b1 #d1 #f1 #h1 #aj1 #ab1 #ad #aje moves4 ,in the odd cases1 write the even numbers in ascending order and the odd numbers in descending order1 repeat this series 7n - #a7 times1 follow with the even numbers in ascending order 7omitting n - #a71 the odd numbers in descending order 7omitting #a71 and conclude with all the numbers 7odd and even7 in their natural order 7omitting #a and n74 ,thus for #aa frogs3 7#b1 #d1 #f1 #h1 #aj1 #aa1 #i1 #g1 #e1 #c1 #a7 repeated #e times1 #b1 #d1 #f1 #h1 #aa1 #i1 #g1 #e1 #c1 and #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i1 #aj #gc moves4 ,this complete general solution is published here for the first time4 #bae4--,,the ,,grasshopper ,,puzzle4 ,move the counters in the following order4 ,the moves in brackets are to be made four times in succession4 #ab1 #a1 #c1 #b1 #ab1 #aa1 #a1 #c1 #b 7#e1 #g1 #i1 #aj1 #h1 #f1 #d71 #c1 #b1 #ab1 #aa1 #b1 #a1 #b4 ,the grasshoppers will then be reversed in forty-four moves4 ,the general solution of this problem is very difficult4 ,of course it can always be solved by the method given in the solution of the last puzzle1 if we have no desire to use the fewest possible moves4 ,but to employ a full economy of moves we have two main points to consider4 ,there are always what ,i call a lower movement 7,l7 and an upper movement 7,u74 ,l consists in exchanging certain of the highest numbers1 such as #ab1 #aa1 #aj in our ,grasshopper ,puzzle1 with certain of the lower numbers1 #a1 #b1 #c2 the former moving in a clockwise direction1 the latter in a non-clockwise direction4 ,u consists in reversing the intermediate counters4 ,in the above solution for #ab1 it will be seen that #ab1 #aa1 and #a1 #b1 #c are engaged in the ,l movement1 and #d1 #e1 #f1 #g1 #h1 #i1 #aj in the ,u movement4 ,the ,l movement needs #af moves and ,u #bh1 making together #dd4 ,we might also involve #aj in the ,l movement1 which would result in ,l #bc1 ,u #ba1 making also together #dd moves4 ,these ,i call the first and second methods4 ,but any other scheme will entail an increase of moves4 ,you always get these two methods 7of equal economy7 for odd or even counters1 but the point is to determine just how many to involve in ,l and how many in ,u4 ,here is the solution in table form4 ,but first note1 in giving values to n1 that #b1 #c1 and #d counters are special cases1 requiring respectively #c1 #c1 and #f moves1 and that #e and #f counters do not give a minimum solution by the second method--only by the first4 ,,first ,,method4 ---------------------------------------- ------------------------------- ,total ,no4 ,l ,,movement4 ,u ,,movement4 of ---------------------------------------- -------- ,total ,no4 ,counters4 ,no4 of ,no4 of ,no4 of ,no4 of of ,moves4 ,counters4 ,moves4 ,counters4 ,moves4 ---------------------------------------- ----------------------------- #dn n-#a and n #b7n-#a7en-#g #bn#a #bn#cn#a #d7nn-#a7 #dn-#b n-#a n #b7n-#a7en-#g #bn-#a #b7n-#a7cn-#bdn-#e #dn#a n n#a #bn#en-#b #bn #bn#cn-#d #b7bn#dn-#c7 #dn-#a n-#a n #b7n-#a7en-#g #bn #bn#cn-#d #dn#dn-#i ---------------------------------------- ----------------------------- ,,second ,,method4 ---------------------------------------- ------------------------------- ,total ,no4 ,l ,,movement4 ,u ,,movement4 of ---------------------------------------- --------- ,total ,no4 ,counters4 ,no4 of ,no4 of ,no4 of ,no4 of of ,moves4 ,counters4 ,moves4 ,counters4 ,moves4 ---------------------------------------- ----------------------------- #dn n and n #bn#cn-#d #bn #b7n-#a7en-#b #d7nn-#a7 #dn-#b n-#a n-#a #b7n-#a7cn-#g #bn #b7n-#a7en-#b #dn-#e #dn#a n n #bn#cn-#d #bn#a #bn#en-#b #b7bn#dn-#c7 #dn-#a n n #bn#cn-#d #bn-#a #b7n-#a7en-#g #dn#dn-#i ---------------------------------------- ----------------------------- ,more generally we may say that with m counters1 where m is even and greater than #d1 we require 7m #dm - #af7_/d moves2 and where m is odd and greater than #c1 7m #fm - #ca7_/d moves4 ,i have thus shown the reader how to find the minimum number of moves for any case1 and the character and direction of the moves4 ,i will leave him to discover for himself how the actual order of moves is to be determined4 ,this is a hard nut1 and requires careful adjustment of the ,l and the ,u movements1 so that they may be mutually accommodating4 #baf4--,,the ,,educated ,,frogs4 ,the following leaps solve the puzzle in ten moves3 #b to #a1 #e to #b1 #c to #e1 #f to #c1 #g to #f1 #d to #g1 #a to #d1 #c to #a1 #f to #c1 #g to #f4 #bag4--,,the ,,twickenham ,,puzzle4 ,play the counters in the following order3 ,k ,c ,e ,k ,w ,t ,c ,e ,h ,m ,k ,w ,t ,a ,n ,c ,e ,h ,m ,i ,k ,c ,e ,h ,m ,t1 and there you are1 at ,twickenham4 ,the position itself will always determine whether you are to make a leap or a simple move4 #bah4--,,the ,,victoria ,,cross ,,puzzle4 ,in solving this puzzle there were two things to be achieved3 first1 so to manipulate the counters that the word ,,victoria should read round the cross in the same direction1 only with the ,v on one of the dark arms2 and secondly1 to perform the feat in the fewest possible moves4 ,now1 as a matter of fact1 it would be impossible to perform the first part in any way whatever if all the letters of the word were different2 but as there are two ,i's1 it can be done by making these letters change places--that is1 the first ,i changes from the #bnd place to the #gth1 and the second ,i from the #gth place to the #bnd4 ,but the point ,i referred to1 when introducing the puzzle1 as a little remarkable is this3 that a solution in twenty-two moves is obtainable by moving the letters in the order of the following words3 ,a ,,victor6 ,a ,,victor6 ,a ,,victor ,i6 ,there are1 however1 just six solutions in eighteen moves1 and the following is one of them3 ,i 7#a71 ,v1 ,a1 ,i 7#b71 ,r1 ,o1 ,t1 ,i 7#a71 ,i 7#b71 ,a1 ,v1 ,i 7#b71 ,i 7#a71 ,c1 ,i 7#b71 ,v1 ,a1 ,i 7#a74 ,the first and second ,i in the word are distinguished by the numbers #a and #b4 ,it will be noticed that in the first solution given above one of the ,i's never moves1 though the movements of the other letters cause it to change its relative position4 ,there is another peculiarity ,i may point out--that there is a solution in twenty-eight moves requiring no letter to move to the central division except the ,i's4 ,i may also mention that1 in each of the solutions in eighteen moves1 the letters ,c1 ,t1 ,o1 ,r move once only1 while the second ,i always moves four times1 the ,v always being transferred to the right arm of the cross4 #bai4--,,the ,,letter ,,block ,,puzzle4 ,this puzzle can be solved in #bc moves--the fewest possible4 ,move the blocks in the following order3 ,a1 ,b1 ,f1 ,e1 ,c1 ,a1 ,b1 ,f1 ,e1 ,c1 ,a1 ,b1 ,d1 ,h1 ,g1 ,a1 ,b1 ,d1 ,h1 ,g1 ,d1 ,e1 ,f4 #bbj4--;,a ,,lodging-,,house ,,difficulty4 ,the shortest possible way is to move the articles in the following order3 ,piano1 bookcase1 wardrobe1 piano1 cabinet1 chest of drawers1 piano1 wardrobe1 bookcase1 cabinet1 wardrobe1 piano1 chest of drawers1 wardrobe1 cabinet1 bookcase1 piano4 ,thus seventeen removals are necessary4 ,the landlady could then move chest of drawers1 wardrobe1 and cabinet4 ,mr4 ,dobson did not mind the wardrobe and chest of drawers changing rooms so long as he secured the piano4 #bba4--,,the ,,eight ,,engines4 ,the solution to the ,eight ,engines ,puzzle is as follows3 ,the engine that has had its fire drawn and therefore cannot move is ,no4 #e4 ,move the other engines in the following order3 #g1 #f1 #c1 #g1 #f1 #a1 #b1 #d1 #a1 #c1 #h1 #a1 #c1 #b1 #d1 #c1 #b1 seventeen moves in all1 leaving the eight engines in the required order4 ,there are two other slightly different solutions4 #bbb4--;,a ,,railway ,,puzzle4 ,this little puzzle may be solved in as few as nine moves4 ,play the engines as follows3 ,from #i to #aj1 from #f to #i1 from #e to #f1 from #b to #e1 from #a to #b1 from #g to #a1 from #h to #g1 from #i to #h1 and from #aj to #i4 ,you will then have engines ,a1 ,b1 and ,c on each of the three circles and on each of the three straight lines4 ,this is the shortest solution that is possible4 #bbc4--;,a ,,railway ,,muddle4 ,illustration3 #a ,illustration3 #b ,illustration3 #c ,illustration3 #d ,illustration3 #e ,illustration3 #f ,only six reversals are necessary4 ,the white train 7from ,a to ,d7 is divided into three sections1 engine and #g wagons1 #h wagons1 and #a wagon4 ,the black train 7,d to ,a7 never uncouples anything throughout4 ,fig4 #a is original position with #h and #a uncoupled4 ,the black train proceeds to position in ,fig4 #b 7no reversal74 ,the engine and #g proceed towards ,d1 and black train backs1 leaves #h on loop1 and takes up position in ,fig4 #c 7first reversal74 ,black train goes to position in ,fig4 #d to fetch single wagon 7second reversal74 ,black train pushes #h off loop and leaves single wagon there1 proceeding on its journey1 as in ,fig4 #e 7third and fourth reversals74 ,white train now backs on to loop to pick up single car and goes right away to ,d 7fifth and sixth reversals74 #bbd4--,,the ,,motor-,,garage ,,puzzle4 ,the exchange of cars can be made in forty-three moves1 as follows3 #f-;,g1 #b-;,b1 #a-;,e1 #c-;,h1 #d-;,i1 #c-,l1 #f-,k1 #d-;,g1 #a-;,i1 #b-;,j1 #e-;,h1 #d-;,a1 #g-;,f1 #h-;,e1 #d-;,d1 #h-;,c1 #g-;,a1 #h-;,g1 #e-;,c1 #b-;,b1 #a-;,e1 #h-;,i1 #a-;,g1 #b-;,j1 #g-;,h1 #a-;,a1 #g-;,g1 #b-;,b1 #f-;,e1 #c-;,h1 #h-,l1 #c-;,i1 #g-,k1 #c-;,g1 #f-;,i1 #b-;,j1 #e-;,h1 #c-;,c1 #e-;,g1 #b-;,b1 #f-;,e1 #e-;,i1 #f-;,j4 ,of course1 #f-;,g means that the car numbered #f moves to the point ,g4 ,there are other ways in forty-three moves4 #bbe4--,,the ,,ten ,,prisoners4 ,illustration ,it will be seen in the illustration how the prisoners may be arranged so as to produce as many as sixteen even rows4 ,there are #d such vertical rows1 #d horizontal rows1 #e diagonal rows in one direction1 and #c diagonal rows in the other direction4 ,the arrows here show the movements of the four prisoners1 and it will be seen that the infirm man in the bottom corner has not been moved4 #bbf4--,,round ,,the ,,coast4 ,in order to place words round the circle under the conditions1 it is necessary to select words in which letters are repeated in certain relative positions4 ,thus1 the word that solves our puzzle is ,swansea1 in which the first and fifth letters are the same1 and the third and seventh the same4 ,we make out jumps as follows1 taking the letters of the word in their proper order3 #b-e1 #g-b1 #d-g1 #a-d1 #f-a1 #c-f1 #h-c4 ,or we could place a word like ,tarapur 7in which the second and fourth letters1 and the third and seventh1 are alike7 with these moves3 #f-a1 #g-d1 #b-g1 #e--b1 #h-e1 #c-f1 #h-c4 ,but ,swansea is the only word1 apparently1 that will fulfil the conditions of the puzzle4 ,this puzzle should be compared with ,sharp's ,puzzle1 referred to in my solution to ,no4 #cda1 ,the ,four ,frogs4 ,the condition touch and jump over two is identical with touch and move along a line4 #bbg4--;,,central ,,solitaire4 ,here is a solution in nineteen moves2 the moves enclosed in brackets count as one move only3 #ai-ag1 #af-ah1 7#bi-ag1 #ag-ai71 #cj-ah1 #bg-be1 7#bb-bd1 #bd-bf71 #ca-bc1 7#d-af1 #af-bh71 #g-i1 #aj-h1 #ab-aj1 #c-aa1 #ah-f1 7#a-c1 #c-aa71 7#ac-bg1 #bg-be71 7#ba-g1 #g-i71 7#cc-ca1 #ca-bc71 7#aj-h1 #h-bb1 #bb-bd1 #bd-bf1 #bf-ab1 #ab-aj71 #e-ag4 ,all the counters are now removed except one1 which is left in the central hole4 ,the solution needs judgment1 as one is tempted to make several jumps in one move1 where it would be the reverse of good play4 ,for example1 after playing the first #c-aa above1 one is inclined to increase the length of the move by continuing with #aa-be1 #be-bg1 or with #aa-i1 #i-g4 ,i do not think the number of moves can be reduced4 #bbh4--,,the ,,ten ,,apples4 ,number the plates 7#a1 #b1 #c1 #d71 7#e1 #f1 #g1 #h71 7#i1 #aj1 #aa1 #ab71 7#ac1 #ad1 #ae1 #af7 in successive rows from the top to the bottom4 ,then transfer the apple from #h to #aj and play as follows1 always removing the apple jumped over3 #i-aa1 #a-i1 #ac-e1 #af-h1 #d-ab1 #ab-aj1 #c-a1 #a-i1 #i-aa4 #bbi4--,,the ,,nine ,,almonds4 ,this puzzle may be solved in as few as four moves1 in the following manner3 ,move #e over #h1 #i1 #c1 #a4 ,move #g over #d4 ,move #f over #b and #g4 ,move #e over #f1 and all the counters are removed except #e1 which is left in the central square that it originally occupied4 #bcj4--,,the ,,twelve ,,pennies4 ,here is one of several solutions4 ,move #ab to #c1 #g to #d1 #aj to #f1 #h to #a1 #i to #e1 #aa to #b4 #bca4--,,plates ,,and ,,coins4 ,number the plates from #a to #ab in the order that the boy is seen to be going in the illustration4 ,starting from #a1 proceed as follows1 where #a to #d means that you take the coin from plate ,no4 #a and transfer it to plate ,no4 #d3 #a to #d1 #e to #h1 #i to #ab1 #c to #f1 #g to #aj1 #aa to #b1 and complete the last revolution to #a1 making three revolutions in all4 ,or you can proceed this way3 #d to #g1 #h to #aa1 #ab to #c1 #b to #e1 #f to #i1 #aj to #a4 ,it is easy to solve in four revolutions1 but the solutions in three are more difficult to discover4 ,this is ,the ,riddle of the ,fishpond 7,no4 #da1 ,canterbury ,puzzles7 in a different dress4 #bcb4--;,,catching ,,the ,,mice4 ,in order that the cat should eat every thirteenth mouse1 and the white mouse last of all1 it is necessary that the count should begin at the seventh mouse 7calling the white one the first7--that is1 at the one nearest the tip of the cat's tail4 ,in this case it is not at all necessary to try starting at all the mice in turn until you come to the right one1 for you can just start anywhere and note how far distant the last one eaten is from the starting point4 ,you will find it to be the eighth1 and therefore must start at the eighth1 counting backwards from the white mouse4 ,this is the one ,i have indicated4 ,in the case of the second puzzle1 where you have to find the smallest number with which the cat may start at the white mouse and eat this one last of all1 unless you have mastered the general solution of the problem1 which is very difficult1 there is no better course open to you than to try every number in succession until you come to one that works correctly4 ,the smallest number is twenty-one4 ,if you have to proceed by trial1 you will shorten your labour a great deal by only counting out the remainders when the number is divided successively by #ac1 #ab1 #aa1 #aj1 etc4 ,thus1 in the case of #ba1 we have the remainders #h1 #i1 #aj1 #a1 #c1 #e1 #g1 #c1 #a1 #a1 #c1 #a1 #a4 ,note that ,i do not give the remainders of #g1 #c1 and #a as nought1 but as #g1 #c1 and #a4 ,now1 count round each of these numbers in turn1 and you will find that the white mouse is killed last of all4 ,of course1 if we wanted simply any number1 not the smallest1 the solution is very easy1 for we merely take the least common multiple of #ac1 #ab1 #aa1 #aj1 etc4 down to #b4 ,this is #cfjcfj1 and you will find that the first count kills the thirteenth mouse1 the next the twelfth1 the next the eleventh1 and so on down to the first4 ,but the most arithmetically inclined cat could not be expected to take such a big number when a small one like twenty-one would equally serve its purpose4 ,in the third case1 the smallest number is #ajj4 ,the number #a1jjj would also do1 and there are just seventy-two other numbers between these that the cat might employ with equal success4 #bcc4--,,the ,,eccentric ,,cheesemonger4 ,to leave the three piles at the extreme ends of the rows1 the cheeses may be moved as follows--the numbers refer to the cheeses and not to their positions in the row3 #g-b1 #h-g1 #i-h1 #aj-ae1 #f-aj1 #e-f1 #ad-af1 #ac-ad1 #ab-ac1 #c-a1 #d-c1 #aa-d4 ,this is probably the easiest solution of all to find4 ,to get three of the piles on cheeses #ac1 #ad1 and #ae1 play thus3 #i-d1 #aj-i1 #aa-aj1 #f-ad1 #e-f1 #ab-ae1 #h-ab1 #g-h1 #af-e1 #c-ac1 #b-c1 #a-b4 ,to leave the piles on cheeses #c1 #e1 #ab1 and #ad1 play thus3 #h-c1 #i-ad1 #af-ab1 #a-e1 #aj-i1 #g-aj1 #aa-h1 #b-a1 #d-af1 #ac-b1 #f-aa1 #ae-d4 #bcd4--,,the ,,exchange ,,puzzle4 ,make the following exchanges of pairs3 ,h-,k1 ,h-,e1 ,h-,c1 ,h-,a1 ,i-,l1 ,i-,f1 ,i-,d1 ,k-,l1 ,g-,j1 ,j-,a1 ,f-,k1 ,l-,e1 ,d-,k1 ,e-,f1 ,e-,d1 ,e-,b1 ,b-,k4 ,it will be found that1 although the white counters can be moved to their proper places in #aa moves1 if we omit all consideration of exchanges1 yet the black cannot be so moved in fewer than #ag moves4 ,so we have to introduce waste moves with the white counters to equal the minimum required by the black4 ,thus fewer than #ag moves must be impossible4 ,some of the moves are1 of course1 interchangeable4 #bce4--,,torpedo ,,practice4 ,illustration3 #aj #f #g _/ #d u u #b u _/ #c-u u u u u u u u u u -----#i--- _/ u #h u u _/ #a #e ,if the enemy's fleet be anchored in the formation shown in the illustration1 it will be seen that as many as ten out of the sixteen ships may be blown up by discharging the torpedoes in the order indicated by the numbers and in the directions indicated by the arrows4 ,as each torpedo in succession passes under three ships and sinks the fourth1 strike out each vessel with the pencil as it is sunk4 #bcf4--,,the ,,hat ,,puzzle4 ,illustration3 #a #b #c #d #e #f #g #h #i #aj #aa #ab ------------------------ 99 o 99 ,o 99 ,o 99 ,o 99 ,o ------------------------ 99 ,o 99 ,o 99 ,o 99 ,o ,o 99 ------------------------ 99 99 ,o ,o 99 ,o 99 ,o ,o 99 ------------------------ 99 99 ,o ,o ,o 99 99 ,o ,o 99 ------------------------ 99 99 ,o ,o ,o ,o ,o 99 99 99 ------------------------ ,o ,o ,o ,o ,o 99 99 99 99 99 ------------------------ ,i suggested that the reader should try this puzzle with counters1 so ,i give my solution in that form4 ,the silk hats are represented by black counters and the felt hats by white counters4 ,the first row shows the hats in their original positions1 and then each successive row shows how they appear after one of the five manipulations4 ,it will thus be seen that we first move hats #b and #c1 then #g and #h1 then #d and #e1 then #aj and #aa1 and1 finally1 #a and #b1 leaving the four silk hats together1 the four felt hats together1 and the two vacant pegs at one end of the row4 ,the first three pairs moved are dissimilar hats1 the last two pairs being similar4 ,there are other ways of solving the puzzle4 #bcg4--;,,boys ,,and ,,girls4 ,there are a good many different solutions to this puzzle4 ,any contiguous pair1 except #g-h1 may be moved first1 and after the first move there are variations4 ,the following solution shows the position from the start right through each successive move to the end3-- 4 4 #a #b #c #d #e #f #g #h #d #c #a #b 4 4 #e #f #g #h #d #c #a #b #g #f #e 4 4 #h #d #c #a #b #g 4 4 #e #f #h #d 4 4 #b #g #a #c #e #f #h #d #h #f #b #g #a #c #e 4 4 #bch4--;,,arranging ,,the ,,jam ,,pots4 ,two of the pots1 #ac and #ai1 were in their proper places4 ,as every interchange may result in a pot being put in its place1 it is clear that twenty-two interchanges will get them all in order4 ,but this number of moves is not the fewest possible1 the correct answer being seventeen4 ,exchange the following pairs3 7#c-a1 #b-c71 7#ae-d1 #af-ae71 7#ag-g1 #bj-ag71 7#bd-aj1 #aa-bd1 #ab-aa71 7#h-e1 #f-h1 #ba-f1 #bc-ba1 #bb-bc1 #ad-bb1 #i-ad1 #ah-i74 ,when you have made the interchanges within any pair of brackets1 all numbers within those brackets are in their places4 ,there are five pairs of brackets1 and #e from #bb gives the number of changes required--#ag4 #bci4--;,a ,,juvenile ,,puzzle4 ,illustration3 ----------------- ,c ,e ,d ,f ---------------,b ,g ,a ,h ----------------- ,as the conditions are generally understood1 this puzzle is incapable of solution4 ,this can be demonstrated quite easily4 ,so we have to look for some catch or quibble in the statement of what we are asked to do4 ,now if you fold the paper and then push the point of your pencil down between the fold1 you can with one stroke make the two lines ,,cd and ,,ef in our diagram4 ,then start at ,a1 and describe the line ending at ,b4 ,finally put in the last line ,,gh1 and the thing is done strictly within the conditions1 since folding the paper is not actually forbidden4 ,of course the lines are here left unjoined for the purpose of clearness4 ,in the rubbing out form of the puzzle1 first rub out ,a to ,b with a single finger in one stroke4 ,then rub out the line ,,gh with one finger4 ,finally1 rub out the remaining two vertical lines with two fingers at once6 ,that is the old trick4 #bdj4--,,the ,,union ,,jack4 ,illustration3 ------- ----- ,a ,b _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ -------_/--_/----- _/ _/ _/ _/ _/ _/ _/ -----_/--_/------- _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ ----- ----- ,there are just sixteen points 7all on the outside7 where three roads may be said to join4 ,these are called by mathematicians odd nodes4 ,there is a rule that tells us that in the case of a drawing like the present one1 where there are sixteen odd nodes1 it requires eight separate strokes or routes 7that is1 half as many as there are odd nodes7 to complete it4 ,as we have to produce as much as possible with only one of these eight strokes1 it is clearly necessary to contrive that the seven strokes from odd node to odd node shall be as short as possible4 ,start at ,a and end at ,b1 or go the reverse way4 #bda4--,,the ,,dissected ,,circle4 ,illustration3 _/--------------- _/ _/ _/------,b _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ ,a _/ _/ _/ _/ _/ ----------99---------- _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ ,d-------99----------99----,e _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ ----------99---------- _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ _/ ------------_/ _/ _/ ,c-------_/ ,it can be done in twelve continuous strokes1 thus3 ,start at ,a in the illustration1 and eight strokes1 forming the star1 will bring you back to ,a2 then one stroke round the circle to ,b1 one stroke to ,c1 one round the circle to ,d1 and one final stroke to ,e--twelve in all4 ,of course1 in practice the second circular stroke will be over the first one2 it is separated in the diagram1 and the points of the star not joined to the circle1 to make the solution clear to the eye4 #bdb4--,,the ,,tube ,,inspector',s ,,puzzle4 ,the inspector need only travel nineteen miles if he starts at ,b and takes the following route3 ,,badgdefifcbehklihgjk4 ,thus the only portions of line travelled over twice are the two sections ,d to ,g and ,f to ,i4 ,of course1 the route may be varied1 but it cannot be shortened4 #bdc4--,,visiting ,,the ,,towns4 ,note that there are six towns1 from which only two roads issue4 ,thus #a must lie between #i and #ab in the circular route4 ,mark these two roads as settled4 ,similarly mark #i1 #e1 #ad1 and #d1 #h1 #ad1 and #aj1 #f1 #ae1 and #aj1 #b1 #ac1 and #c1 #g1 #ac4 ,all these roads must be taken4 ,then you will find that he must go from #d to #ae1 as #ac is closed1 and that he is compelled to take #c1 #aa1 #af1 and also #af1 #ab4 ,thus1 there is only one route1 as follows3 #a1 #i1 #e1 #ad1 #h1 #d1 #ae1 #f1 #aj1 #b1 #ac1 #g1 #c1 #aa1 #af1 #ab1 #a1 or its reverse--reading the line the other way4 ,seven roads are not used4 #bdd4--,,the ,,fifteen ,,turnings4 ,illustration ,it will be seen from the illustration 7where the roads not used are omitted7 that the traveller can go as far as seventy miles in fifteen turnings4 ,the turnings are all numbered in the order in which they are taken4 ,it will be seen that he never visits nineteen of the towns4 ,he might visit them all in fifteen turnings1 never entering any town twice1 and end at the black town from which he starts 7see ,the ,rook's ,tour1 ,no4 #cbj71 but such a tour would only take him sixty-four miles4 #bde4--,,the ,,fly ,,on ,,the ,,octahedron4 ,illustration ,though we cannot really see all the sides of the octahedron at once1 we can make a projection of it that suits our purpose just as well4 ,in the diagram the six points represent the six angles of the octahedron1 and four lines proceed from every point under exactly the same conditions as the twelve edges of the solid4 ,therefore if we start at the point ,a and go over all the lines once1 we must always end our route at ,a4 ,and the number of different routes is just #a1dhh1 counting the reverse way of any route as different4 ,it would take too much space to show how ,i make the count4 ,it can be done in about five minutes1 but an explanation of the method is difficult4 ,the reader is therefore asked to accept my answer as correct4 #bdf4--,,the ,,icosahedron ,,puzzle4 ,illustration ,there are thirty edges1 of which eighteen were visible in the original illustration1 represented in the following diagram by the hexagon ,,naesgd4 ,by this projection of the solid we get an imaginary view of the remaining twelve edges1 and are able to see at once their direction and the twelve points at which all the edges meet4 ,the difference in the length of the lines is of no importance2 all we want is to present their direction in a graphic manner4 ,but in case the novice should be puzzled at only finding nineteen triangles instead of the required twenty1 ,i will point out that the apparently missing triangle is the outline ,,hik4 ,in this case there are twelve odd nodes2 therefore six distinct and disconnected routes will be needful if we are not to go over any lines twice4 ,let us therefore find the greatest distance that we may so travel in one route4 ,it will be noticed that ,i have struck out with little cross strokes five lines or edges in the diagram4 ,these five lines may be struck out anywhere so long as they do not join one another1 and so long as one of them does not connect with ,n1 the ,north ,pole1 from which we are to start4 ,it will be seen that the result of striking out these five lines is that all the nodes are now even except ,n and ,s4 ,consequently if we begin at ,n and stop at ,s we may go over all the lines1 except the five crossed out1 without traversing any line twice4 ,there are many ways of doing this4 ,here is one route3 ,n to ,h1 ,i1 ,k1 ,s1 ,i1 ,e1 ,s1 ,g1 ,k1 ,d1 ,h1 ,a1 ,n1 ,b1 ,a1 ,e1 ,f1 ,b1 ,c1 ,g1 ,d1 ,n1 ,c1 ,f1 ,s4 ,by thus making five of the routes as short as is possible--simply from one node to the next--we are able to get the greatest possible length for our sixth line4 ,a greater distance in one route1 without going over the same ground twice1 it is not possible to get4 ,it is now readily seen that those five erased lines must be gone over twice1 and they may be picked up1 so to speak1 at any points of our route4 ,thus1 whenever the traveller happens to be at ,i he can run up to ,a and back before proceeding on his route1 or he may wait until he is at ,a and then run down to ,i and back to ,a4 ,and so with the other lines that have to be traced twice4 ,it is1 therefore1 clear that he can go over #be of the lines once only 7#be #aj1jjj miles #bej1jjj miles7 and #e of the lines twice 7#e #bj1jjj miles #ajj1jjj miles71 the total1 #cej1jjj miles1 being the length of his travels and the shortest distance that is possible in visiting the whole body4 ,it will be noticed that ,i have made him end his travels at ,s1 the ,south ,pole1 but this is not imperative4 ,i might have made him finish at any of the other nodes1 except the one from which he started4 ,suppose it had been required to bring him home again to ,n at the end of his travels4 ,then instead of suppressing the line ,,ai we might leave that open and close ,,is4 ,this would enable him to complete his #cej1jjj miles tour at ,a1 and another #aj1jjj miles would take him to his own fireside4 ,there are a great many different routes1 but as the lengths of the edges are all alike1 one course is as good as another4 ,to make the complete #cej1jjj miles tour from ,n to ,s absolutely clear to everybody1 ,i will give it entire3 ,n to ,h1 ,i1 ,a1 ,i1 ,k1 ,h1 ,k1 ,s1 ,i1 ,e1 ,s1 ,g1 ,f1 ,g1 ,k1 ,d1 ,c1 ,d1 ,h1 ,a1 ,n1 ,b1 ,e1 ,b1 ,a1 ,e1 ,f1 ,b1 ,c1 ,g1 ,d1 ,n1 ,c1 ,f1 ,s--that is1 thirty-five lines of #aj1jjj miles each4 #bdg4--;,,inspecting ,a ,,mine4 ,starting from ,a1 the inspector need only travel #cf furlongs if he takes the following route3 ,a to ,b1 ,g1 ,h1 ,c1 ,d1 ,i1 ,h1 ,m1 ,n1 ,i1 ,j1 ,o1 ,n1 ,s1 ,r1 ,m1 ,l1 ,g1 ,f1 ,k1 ,l1 ,q1 ,r1 ,s1 ,t1 ,o1 ,j1 ,e1 ,d1 ,c1 ,b1 ,a1 ,f1 ,k1 ,p1 ,q4 ,he thus passes between ,a and ,b twice1 between ,c and ,d twice1 between ,f and ,k twice1 between ,j and ,o twice1 and between ,r and ,s twice--five repetitions4 ,therefore #ca passages plus #e repeated equal #cf furlongs4 ,the little pitfall in this puzzle lies in the fact that we start from an even node4 ,otherwise we need only travel #ce furlongs4 #bdh4--,,the ,,cyclist',s ,,tour4 ,when ,mr4 ,maggs replied1 ,no way1 ,i'm sure1 he was not saying that the thing was impossible1 but was really giving the actual route by which the problem can be solved4 ,starting from the star1 if you visit the towns in the order1 ,,no ,,way1 ,i',m ,,sure1 you will visit every town once1 and only once1 and end at ,e4 ,so both men were correct4 ,this was the little joke of the puzzle1 which is not by any means difficult4 #bdi4--,,the ,,sailor',s ,,puzzle4 ,illustration ,there are only four different routes 7or eight1 if we count the reverse ways7 by which the sailor can start at the island marked ,a1 visit all the islands once1 and once only1 and return again to ,a4 ,here they are3-- ,a ,i ,p ,t ,l ,o ,e ,h ,r ,q ,d ,c ,f ,u ,g ,n ,s ,k ,m ,b ,a ,a ,i ,p ,t ,s ,n ,g ,l ,o ,e ,u ,f ,c ,d ,k ,m ,b ,q ,r ,h ,a ,a ,b ,m ,k ,s ,n ,g ,l ,t ,p ,i ,o ,e ,u ,f ,c ,d ,q ,r ,h ,a ,a ,i ,p ,t ,l ,o ,e ,u ,g ,n ,s ,k ,m ,b ,q ,d ,c ,f ,r ,h ,a ,now1 if the sailor takes the first route he will make ,c his #abth island 7counting ,a as #a72 by the second route he will make ,c his #acth island2 by the third route1 his #afth island2 and by the fourth route1 his #agth island4 ,if he goes the reverse way1 ,c will be respectively his #ajth1 #ith1 #fth1 and #eth island4 ,as these are the only possible routes1 it is evident that if the sailor puts off his visit to ,c as long as possible1 he must take the last route reading from left to right4 ,this route ,i show by the dark lines in the diagram1 and it is the correct answer to the puzzle4 ,the map may be greatly simplified by the buttons and string method1 explained in the solution to ,no4 #cda1 ,the ,four ,frogs4 #bej4--,,the ,,grand ,,tour4 ,the first thing to do in trying to solve a puzzle like this is to attempt to simplify it4 ,if you look at ,fig4 #a1 you will see that it is a simplified version of the map4 ,imagine the circular towns to be buttons and the railways to be connecting strings4 7,see solution to ,no4 #cda47 ,then1 it will be seen1 we have simply straightened out the previous diagram without affecting the conditions4 ,now we can further simplify by converting ,fig4 #a into ,fig4 #b1 which is a portion of a chessboard4 ,here the directions of the railways will resemble the moves of a rook in chess--that is1 we may move in any direction parallel to the sides of the diagram1 but not diagonally4 ,therefore the first town 7or square7 visited must be a black one2 the second must be a white2 the third must be a black2 and so on4 ,every odd square visited will thus be black and every even one white4 ,now1 we have #bc squares to visit 7an odd number71 so the last square visited must be black4 ,but ,z happens to be white1 so the puzzle would seem to be impossible of solution4 ,illustration3 ,fig4 #a4 ,illustration3 ,fig4 #b4 ,as we were told that the man succeeded in carrying put his plan1 we must try to find some loophole in the conditions4 ,he was to enter every town once and only once1 and we find no prohibition against his entering once the town ,a after leaving it1 especially as he has never left it since he was born1 and would thus be entering it for the first time in his life4 ,but he must return at once from the first town he visits1 and then he will have only #bb towns to visit1 and as #bb is an even number1 there is no reason why he should not end on the white square ,z4 ,a possible route for him is indicated by the dotted line from ,a to ,z4 ,this route is repeated by the dark lines in ,fig4 #a1 and the reader will now have no difficulty in applying2 it to the original map4 ,we have thus proved that the puzzle can only be solved by a return to ,a immediately after leaving it4 #bea4--,,water1 ,,gas1 ,,and ,,electricity4 ,illustration ,according to the conditions1 in the strict sense in which one at first understands them1 there is no possible solution to this puzzle4 ,in such a dilemma one always has to look for some verbal quibble or trick4 ,if the owner of house ,a will allow the water company to run their pipe for house ,c through his property 7and we are not bound to assume that he would object71 then the difficulty is got over1 as shown in our illustration4 ,it will be seen that the dotted line from ,w to ,c passes through house ,a1 but no pipe ever crosses another pipe4 #beb4--;,a ,,puzzle ,,for ,,motorists4 ,illustration ,the routes taken by the eight drivers are shown in the illustration1 where the dotted line roads are omitted to make the paths clearer to the eye4 #bec4--;,a ,,bank ,,holiday ,,puzzle4 ,the simplest way is to write in the number of routes to all the towns in this manner4 ,put a #a on all the towns in the top row and in the first column4 ,then the number of routes to any town will be the sum of the routes to the town immediately above and to the town immediately to the left4 ,thus the routes in the second row will be #a1 #b1 #c1 #d1 #e1 #f1 etc41 in the third row1 #a1 #c1 #f1 #aj1 #ae1 #ba1 etc42 and so on with the other rows4 ,it will then be seen that the only town to which there are exactly #a1cfe different routes is the twelfth town in the fifth row--the one immediately over the letter ,e4 ,this town was therefore the cyclist's destination4 ,the general formula for the number of routes from one corner to the corner diagonally opposite on any such rectangular reticulated arrangement1 under the conditions as to direction1 is 7mn76_/m6n61 where m is the number of towns on one side1 less one1 and n the number on the other side1 less one4 ,our solution involves the case where there are #ab towns by #e4 ,therefore m #aa and n #d4 ,then the formula gives us the answer #a1cfe as above4 #bed4-- ,,the ,,motor-,,car ,,tour4 ,first of all ,i will ask the reader to compare the original square diagram with the circular one shown in ,figs4 #a1 #b1 and #c below4 ,if for the moment we ignore the shading 7the purpose of which ,i shall proceed to explain71 we find that the circular diagram in each case is merely a simplification of the original square one--that is1 the roads from ,a lead to ,b1 ,e1 and ,m in both cases1 the roads from ,l 7,london7 lead to ,i1 ,k1 and ,s1 and so on4 ,the form below1 being circular and symmetrical1 answers my purpose better in applying a mechanical solution1 and ,i therefore adopt it without altering in any way the conditions of the puzzle4 ,if such a question as distances from town to town came into the problem1 the new diagrams might require the addition of numbers to indicate these distances1 or they might conceivably not be at all practicable4 ,illustration3 ,figs4 #a1 #b1 and #c ,now1 ,i draw the three circular diagrams1 as shown1 on a sheet of paper and then cut out three pieces of cardboard of the forms indicated by the shaded parts of these diagrams4 ,it can be shown that every route1 if marked out with a red pencil1 will form one or other of the designs indicated by the edges of the cards1 or a reflection thereof4 ,let us direct our attention to ,fig4 #a4 ,here the card is so placed that the star is at the town ,t2 it therefore gives us 7by following the edge of the card7 one of the circular routes from ,london3 ,l1 ,s1 ,r1 ,t1 ,m1 ,a1 ,e1 ,p1 ,o1 ,j1 ,d1 ,c1 ,b1 ,g1 ,n1 ,q1 ,k1 ,h1 ,f1 ,i1 ,l4 ,if we went the other way1 we should get ,l1 ,i1 ,f1 ,h1 ,k1 ,q1 etc41 but these reverse routes were not to be counted4 ,when we have written out this first route we revolve the card until the star is at ,m1 when we get another different route1 at ,a a third route1 at ,e a fourth route1 and at ,p a fifth route4 ,we have thus obtained five different routes by revolving the card as it lies4 ,but it is evident that if we now take up the card and replace it with the other side uppermost1 we shall in the same manner get five other routes by revolution4 ,we therefore see how1 by using the revolving card in ,fig4 #a1 we may1 without any difficulty1 at once write out ten routes4 ,and if we employ the cards in ,figs4 #b and #c1 we similarly obtain in each case ten other routes4 ,these thirty routes are all that are possible4 ,i do not give the actual proof that the three cards exhaust all the possible cases1 but leave the reader to reason that out for himself4 ,if he works out any route at haphazard1 he will certainly find that it falls into one or other of the three categories4 #bee4--,,the ,,level ,,puzzle4 ,let us confine our attention to the ,l in the top left-hand corner4 ,suppose we go by way of the ,e on the right3 we must then go straight on to the ,v1 from which letter the word may be completed in four ways1 for there are four ,e's available through which we may reach an ,l4 ,there are therefore four ways of reading through the right-hand ,e4 ,it is also clear that there must be the same number of ways through the ,e that is immediately below our starting point4 ,that makes eight4 ,if1 however1 we take the third route through the ,e on the diagonal1 we then have the option of any one of the three ,v's1 by means of each of which we may complete the word in four ways4 ,we can therefore spell ,,level in twelve ways through the diagonal ,e4 ,twelve added to eight gives twenty readings1 all emanating from the ,l in the top left-hand corner2 and as the four corners are equal1 the answer must be four times twenty1 or eighty different ways4 #bef4--,,the ,,diamond ,,puzzle4 ,there are #beb different ways4 ,the general formula is that1 for words of n letters 7not palindromes1 as in the case of the next puzzle71 when grouped in this manner1 there are always #b7n#a7 - #d different readings4 ,this does not allow diagonal readings1 such as you would get if you used instead such a word as ,,digging1 where it would be possible to pass from one ,g to another ,g by a diagonal step4 #beg4--,,the ,,deified ,,puzzle4 ,the correct answer is #a1iib different ways4 ,every ,f is either a corner ,f or a side ,f--standing next to a corner in its own square of ,f's4 ,now1 ,,fied may be read from a corner ,f in #af ways2 therefore ,,deif may be read into a corner ,f also in #af ways2 hence ,,deified may be read through a corner ,f in #af #af #bef ways4 ,consequently1 the four corner ,f's give #d #bef #a1jbd ways4 ,then ,,fied may be read from a side ,f in #aa ways1 and ,,deified therefore in #aba ways4 ,but there are eight side ,f's2 consequently these give together #h #aba #ifh ways4 ,add #ifh to #a1jbd and we get the answer1 #a1iib4 ,in this form the solution will depend on whether the number of letters in the palindrome be odd or even4 ,for example1 if you apply the word ,,nun in precisely the same manner1 you will get #fd different readings2 but if you use the word ,,noon1 you will only get #ef1 because you cannot use the same letter twice in immediate succession 7since you must always pass from one letter to another7 or diagonal readings1 and every reading must involve the use of the central ,n4 ,the reader may like to find for himself the general formula in this case1 which is complex and difficult4 ,i will merely add that for such a case as ,,madam1 dealt with in the same way as ,,deified1 the number of readings is #djj4 #beh4-- ,,the ,,voters' ,,puzzle4 ,,the number of readings here is #fc1ejd1 as in the case of ,,was ,,it ,a ,,rat ,i ,,saw 7,no4 #cj1 ,canterbury ,puzzles74 ,the general formula is that for palindromic sentences containing #bn #a letters there are 7#d7bn -#a77 readings4 #bei4-- ,,hannah',s ,,puzzle4 ,starting from any one of the ,n's1 there are #ag different readings of ,,nah1 or #fh 7#d times #ag7 for the #d ,n's4 ,therefore there are also #fh ways of spelling ,,han4 ,if we were allowed to use the same ,n twice in a spelling1 the answer would be #fh times #fh1 or #d1fbd ways4 ,but the conditions were1 always passing from one letter to another4 ,therefore1 for every one of the #ag ways of spelling ,,han with a particular ,n1 there would be #ea ways 7#c times #ag7 of completing the ,,nah1 or #hfg 7#ag times #ea7 ways for the complete word4 ,hence1 as there are four ,n's to use in ,,han1 the correct solution of the puzzle is #c1dfh 7#d times #hfg7 different ways4 #bfj4--,,the ,,honeycomb ,,puzzle4 ,the required proverb is1 ,there is many a slip 'twixt the cup and the lip4 ,start at the ,t on the outside at the bottom right-hand corner1 pass to the ,h above it1 and the rest is easy4 #bfa4-- ,,the ,,monk ,,and ,,the ,,bridges4 ,illustration ,the problem of the ,bridges may be reduced to the simple diagram shown in illustration4 ,the point ,m represents the ,monk1 the point ,i the ,island1 and the point ,y the ,monastery4 ,now the only direct ways from ,m to ,i are by the bridges a and b2 the only direct ways from ,i to ,y are by the bridges c and d2 and there is a direct way from ,m to ,y by the bridge e4 ,now1 what we have to do is to count all the routes that will lead from ,m to ,y1 passing over all the bridges1 a1 b1 c1 d1 and e once and once only4 ,with the simple diagram under the eye it is quite easy1 without any elaborate rule1 to count these routes methodically4 ,thus1 starting from a1 b1 we find there are only two ways of completing the route2 with a1 c1 there are only two routes2 with a1 d1 only two routes2 and so on4 ,it will be found that there are sixteen such routes in all1 as in the following list3-- a b e c d b c d a e a b e d c b c e a d a c d b e b d c a e a c e b d b d e a c a d e b c e c a b d a d c b e e c b a d b a e c d e d a b c b a e d c e d b a c ,if the reader will transfer the letters indicating the bridges from the diagram to the corresponding bridges in the original illustration1 everything will be quite obvious4 #bfb4--,,those ,,fifteen ,,sheep4 ,if we read the exact words of the writer in the cyclopdia1 we find that we are not told that the pens were all necessarily empty6 ,in fact1 if the reader will refer back to the illustration1 he will see that one sheep is already in one of the pens4 ,it was just at this point that the wily farmer said to me1 ,now ,i'm going to start placing the fifteen sheep4 ,he thereupon proceeded to drive three from his flock into the already occupied pen1 and then placed four sheep in each of the other three pens4 ,there1 says he1 you have seen me place fifteen sheep in four pens so that there shall be the same number of sheep in every pen4 ,i was1 of course1 forced to admit that he was perfectly correct1 according to the exact wording of the question4 #bfc4--,,king ,,arthur',s ,,knights4 ,on the second evening ,king ,arthur arranged the knights and himself in the following order round the table3 ,a1 ,f1 ,b1 ,d1 ,g1 ,e1 ,c4 ,on the third evening they sat thus1 ,a1 ,e1 ,b1 ,g1 ,c1 ,f1 ,d4 ,he thus had ,b next but one to him on both occasions 7the nearest possible71 and ,g was the third from him at both sittings 7the furthest position possible74 ,no other way of sitting the knights would have been so satisfactory4 #bfd4--,,the ,,city ,,luncheons4 ,the men may be grouped as follows1 where each line represents a day and each column a table3-- ,,ab ,,cd ,,ef ,,gh ,,ij ,,kl ,,ae ,,dl ,,gk ,,fi ,,cb ,,hj ,,ag ,,lj ,,fh ,,kc ,,de ,,ib ,,af ,,jb ,,ki ,,hd ,,lg ,,ce ,,ak ,,be ,,hc ,,il ,,jf ,,dg ,,ah ,,eg ,,id ,,cj ,,bk ,,lf ,,ai ,,gf ,,cl ,,db ,,eh ,,jk ,,ac ,,fk ,,dj ,,le ,,gi ,,bh ,,ad ,,kh ,,lb ,,jg ,,fc ,,ei ,,al ,,hi ,,je ,,bf ,,kd ,,gc ,,aj ,,ic ,,bg ,,ek ,,hl ,,fd ,note that in every column 7except in the case of the ,a's7 all the letters descend cyclically in the same order1 ,b1 ,e1 ,g1 ,f1 up to ,j1 which is followed by ,b4 #bfe4--;,a ,,puzzle ,,for ,,card-,,players4 ,in the following solution each of the eleven lines represents a sitting1 each column a table1 and each pair of letters a pair of partners4 ,a ,b -- ,i ,l ,e ,j -- ,g ,k ,f ,h -- ,c ,d ,a ,c -- ,j ,b ,f ,k -- ,h ,l ,g ,i -- ,d ,e ,a ,d -- ,k ,c ,g ,l -- ,i ,b ,h ,j -- ,e ,f ,a ,e -- ,l ,d ,h ,b -- ,j ,c ,i ,k -- ,f ,g ,a ,f -- ,b ,e ,i ,c -- ,k ,d ,j ,l -- ,g ,h ,a ,g -- ,c ,f ,j ,d -- ,l ,e ,k ,b -- ,h ,i ,a ,h -- ,d ,g ,k ,e -- ,b ,f ,l ,c -- ,i ,j ,a ,i -- ,e ,h ,l ,f -- ,c ,g ,b ,d -- ,j ,k ,a ,j -- ,f ,i ,b ,g -- ,d ,h ,c ,e -- ,k ,l ,a ,k -- ,g ,j ,c ,h -- ,e ,i ,d ,f -- ,l ,b ,a ,l -- ,h ,k ,d ,i -- ,f ,j ,e ,g -- ,b ,c ,it will be seen that the letters ,b1 ,c1 ,d 444,l descend cyclically4 ,the solution given above is absolutely perfect in all respects4 ,it will be found that every player has every other player once as his partner and twice as his opponent4 #bff4--;,a ,,tennis ,,tournament4 ,call the men ,a1 ,b1 ,d1 ,e1 and their wives a1 b1 d1 e4 ,then they may play as follows without any person ever playing twice with or against any other person3-- ,first ,court4 ,second ,court4 #ast ,day ,a d against ,b e ,d a against ,e b #bnd ,day ,a e ,d b ,e a ,b d #crd ,day ,a b ,e d ,b a ,d e ,it will be seen that no man ever plays with or against his own wife--an ideal arrangement4 ,if the reader wants a hard puzzle1 let him try to arrange eight married couples 7in four courts on seven days7 under exactly similar conditions4 ,it can be done1 but ,i leave the reader in this case the pleasure of seeking the answer and the general solution4 #bfg4--,,the ,,wrong ,,hats4 ,the number of different ways in which eight persons1 with eight hats1 can each take the wrong hat1 is #ad1hcc4 ,here are the successive solutions for any number of persons from one to eight3-- #a #j #b #a #c #b #d #i #e #dd #f #bfe #g #a1hed #h #ad1hcc ,to get these numbers1 multiply successively by #b1 #c1 #d1 #e1 etc4 ,when the multiplier is even1 add #a2 when odd1 deduct #a4 ,thus1 #c #a - #a #b2 #d #b #a #i2 #e #i - #a #dd2 and so on4 ,or you can multiply the sum of the number of ways for n - #a and n - #b persons by n - #a1 and so get the solution for n persons4 ,thus1 #d7b #i7 #dd2 #e7i #dd7 #bfe2 and so on4 #bfh4--,,the ,,peal ,,of ,,bells4 ,the bells should be rung as follows3-- #a #b #c #d #b #a #d #c #b #d #a #c #d #b #c #a #d #c #b #a #c #d #a #b #c #a #d #b #a #c #b #d #c #a #b #d #a #c #d #b #a #d #c #b #d #a #b #c #d #b #a #c #b #d #c #a #b #c #d #a #c #b #a #d #b #c #a #d #c #b #d #a #c #d #b #a #d #c #a #b #d #a #c #b #a #d #b #c #a #b #d #c #b #a #c #d ,i have constructed peals for five and six bells respectively1 and a solution is possible for any number of bells under the conditions previously stated4 #bfi4--,,three ,,men ,,in ,a ,,boat4 ,if there were no conditions whatever1 except that the men were all to go out together1 in threes1 they could row in an immense number of different ways4 ,if the reader wishes to know how many1 the number is #deeg4 ,and with the condition that no two may ever be together more than once1 there are no fewer than #ae1efg1eeb1jjj different solutions--that is1 different ways of arranging the men4 ,with one solution before him1 the reader will realize why this must be1 for although1 as an example1 ,a must go out once with ,b and once with ,c1 it does not necessarily follow that he must go out with ,c on the same occasion that he goes with ,b4 ,he might take any other letter with him on that occasion1 though the fact of his taking other than ,b would have its effect on the arrangement of the other triplets4 ,of course only a certain number of all these arrangements are available when we have that other condition of using the smallest possible number of boats4 ,as a matter of fact we need employ only ten different boats4 ,here is one the arrangements3-- #a #b #c #d #e #ast ,day 7,,abc7 7,,dbf7 7,,ghi7 7,,jkl7 7,,mno7 #h #f #g #i #aj #bnd ,day 7,,adg7 7,,bkn7 7,,col7 7,,jei7 7,,mhf7 #c #e #d #a #b #crd ,day 7,,ajm7 7,,beh7 7,,cfi7 7,,dko7 7,,gnl7 #g #f #h #i #a #dth ,day 7,,aek7 7,,cgm7 7,,boi7 7,,dhl7 7,,jnf7 #d #e #c #aj #b #eth ,day 7,,ahn7 7,,cdj7 7,,bfl7 7,,geo7 7,,mki7 #f #g #h #aj #a #fth ,day 7,,afo7 7,,bgj7 7,,ckh7 7,,dni7 7,,mel7 #e #d #c #i #b #gth ,day 7,,ail7 7,,bdm7 7,,cen7 7,,gkf7 7,,jho7 ,it will be found that no two men ever go out twice together1 and that no man ever goes out twice in the same boat4 ,this is an extension of the well-known problem of the ,fifteen ,schoolgirls1 by ,kirkman4 ,the original conditions were simply that fifteen girls walked out on seven days in triplets without any girl ever walking twice in a triplet with another girl4 ,attempts at a general solution of this puzzle had exercised the ingenuity of mathematicians since #ahej1 when the question was first propounded1 until recently4 ,in #aijh and the two following years ,i indicated 7see ,educational ,times ,reprints1 ,vols4 ,,xiv41 ,,xv41 and ,,xvii47 that all our trouble had arisen from a failure to discover that #ae is a special case 7too small to enter into the general law for all higher numbers of girls of the form #fn#c71 and showed what that general law is and how the groups should be posed for any number of girls4 ,i gave actual arrangements for numbers that had previously baffled all attempts to manipulate1 and the problem may now be considered generally solved4 ,readers will find an excellent full account of the puzzle in ,w4,w4 ,rouse ,ball's ,mathematical ,recreations1 #eth edition4 #bgj4--,,the ,,glass ,,balls4 ,there are1 in all1 sixteen balls to be broken1 or sixteen places in the order of breaking4 ,call the four strings ,a1 ,b1 ,c1 and ,d--order is here of no importance4 ,the breaking of the balls on ,a may occupy any #d out of these #af places--that is1 the combinations of #af things1 taken #d together1 will be #ac #ad #ae #af ----------------- #a1hbj #a #b #c #d ways for ,a4 ,in every one of these cases ,b may occupy any #d out of the remaining #ab places1 making #i #aj #aa #ab ----------------- #die #a #b #c #d ways4 ,thus #a1hbj #die #ijj1ijj different placings are open to ,a and ,b4 ,but for every one of these cases ,c may occupy #e #f #g #h ------------- #gj #a #b #c #d different places2 so that #ijj1ijj #gj #fc1jfc1jjj different placings are open to ,a1 ,b1 and ,c4 ,in every one of these cases1 ,d has no choice but to take the four places that remain4 ,therefore the correct answer is that the balls may be broken in #fc1jfc1jjj different ways under the conditions4 ,readers should compare this problem with ,no4 #cde1 ,the ,two ,pawns1 which they will then know how to solve for cases where there are three1 four1 or more pawns on the board4 #bga4--;,,fifteen ,,letter ,,puzzle4 ,the following will be found to comply with the conditions of grouping3-- ,,ale ,,met ,,mop ,,blm ,,bag ,,cap ,,you ,,clt ,,ire ,,oil ,,lug ,,lnr ,,nay ,,bit ,,bun ,,bpr ,,aim ,,bey ,,rum ,,gmy ,,oar ,,gin ,,ply ,,cgr ,,peg ,,icy ,,try ,,cmn ,,cue ,,cob ,,tau ,,pnt ,,one ,,got ,,piu ,the fifteen letters used are ,a1 ,e1 ,i1 ,o1 ,u1 ,y1 and ,b1 ,c1 ,g1 ,l1 ,m1 ,n1 ,p1 ,r1 ,t4 ,the number of words is #bg1 and these are all shown in the first three columns4 ,the last word1 ,,piu1 is a musical term in common use2 but although it has crept into some of our dictionaries1 it is ,italian1 meaning a little2 slightly4 ,the remaining twenty-six are good words4 ,of course a ,,tau-cross is a ,t-shaped cross1 also called the cross of ,st4 ,anthony1 and borne on a badge in the ,bishop's ,palace at ,exeter4 ,it is also a name for the toad-fish4 ,we thus have twenty-six good words and one doubtful1 obtained under the required conditions1 and ,i do not think it will be easy to improve on this answer4 ,of course we are not bound by dictionaries but by common usage4 ,if we went by the dictionary only in a case of this kind1 we should find ourselves involved in prefixes1 contractions1 and such absurdities as ,i4,o4,u41 which ,nuttall actually gives as a word4 #bgb4--,,the ,,nine ,,schoolboys4 ,the boys can walk out as follows3-- #ast ,day4 #bnd ,day4 #crd ,day4 ,a ,b ,c ,b ,f ,h ,f ,a ,g ,d ,e ,f ,e ,i ,a ,i ,d ,b ,g ,h ,i ,c ,g ,d ,h ,c ,e #dth ,day4 #eth ,day4 #fth ,day4 ,a ,d ,h ,g ,b ,i ,d ,c ,a ,b ,e ,g ,c ,f ,d ,e ,h ,b ,f ,i ,c ,h ,a ,e ,i ,g ,f ,every boy will then have walked by the side of every other boy once and once only4 ,dealing with the problem generally1 #abn#i boys may walk out in triplets under the conditions on #in#f days1 where n may be nought or any integer4 ,every possible pair will occur once4 ,call the number of boys m4 ,then every boy will pair m-#a times1 of which 7m-#a7_/d times he will be in the middle of a triplet and 7m-#a7_/b times on the outside4 ,thus1 if we refer to the solution above1 we find that every boy is in the middle twice 7making #d pairs7 and four times on the outside 7making the remaining #d pairs of his #h74 ,the reader may now like to try his hand at solving the two next cases of #ba boys on #ae days1 and #cc boys on #bd days4 ,it is1 perhaps1 interesting to note that a school of #dhi boys could thus walk out daily in one leap year1 but it would take #gca girls 7referred to in the solution to ,no4 #bfi7 to perform their particular feat by a daily walk in a year of #cfe days4 #bgc4--,,the ,,round ,,table4 ,the history of this problem will be found in ,the ,canterbury ,puzzles 7,no4 #ij74 ,since the publication of that book in #aijg1 so far as ,i know1 nobody has succeeded in solving the case for that unlucky number of persons1 #ac1 seated at a table on #ff occasions4 ,a solution is possible for any number of persons1 and ,i have recorded schedules for every number up to #be persons inclusive and for #cc4 ,but as ,i know a good many mathematicians are still considering the case of #ac1 ,i will not at this stage rob them of the pleasure of solving it by showing the answer4 ,but ,i will now display the solutions for all the cases up to #ab persons inclusive4 ,some of these solutions are now published for the first time1 and they may afford useful clues to investigators4 ,the solution for the case of #c persons seated on #a occasion needs no remark4 ,a solution for the case of #d persons on #c occasions is as follows3-- #a #b #c #d #a #c #d #b #a #d #b #c ,each line represents the order for a sitting1 and the person represented by the last number in a line must1 of course1 be regarded as sitting next to the first person in the same line1 when placed at the round table4 ,the case of #e persons on #f occasions may be solved as follows3-- #a #b #c #d #e #a #b #d #e #c #a #b #e #c #d --------- #a #c #b #e #d #a #d #b #c #e #a #e #b #d #c ,the case for #f persons on #aj occasions is solved thus3-- #a #b #c #f #d #e #a #c #d #b #e #f #a #d #e #c #f #b #a #e #f #d #b #c #a #f #b #e #c #d ----------- #a #b #d #e #f #c #a #c #e #f #b #d #a #d #f #b #c #e #a #e #b #c #d #f #a #f #c #d #e #b ,it will now no longer be necessary to give the solutions in full1 for reasons that ,i will explain4 ,it will be seen in the examples above that the #a 7and1 in the case of #e persons1 also the #b7 is repeated down the column4 ,such a number ,i call a repeater4 ,the other numbers descend in cyclical order4 ,thus1 for #f persons we get the cycle1 #b1 #c1 #d1 #e1 #f1 #b1 and so on1 in every column4 ,so it is only necessary to give the two lines #a #b #c #f #d #e and #a #b #d #e #f #c1 and denote the cycle and repeaters1 to enable any one to write out the full solution straight away4 ,the reader may wonder why ,i do not start the last solution with the numbers in their natural order1 #a #b #c #d #e #f4 ,if ,i did so the numbers in the descending cycle would not be in their natural order1 and it is more convenient to have a regular cycle than to consider the order in the first line4 ,the difficult case of #g persons on #ae occasions is solved as follows1 and was given by me in ,the ,canterbury ,puzzles3-- #a #b #c #d #e #g #f #a #f #b #g #e #c #d #a #c #e #b #f #g #d #a #e #g #d #c #f #b #a #e #b #g #c #d #f ,in this case the #a is a repeater1 and there are two separate cycles1 #b1 #c1 #d1 #b1 and #e1 #f1 #g1 #e4 ,we thus get five groups of three lines each1 for a fourth line in any group will merely repeat the first line4 ,a solution for #h persons on #ba occasions is as follows3-- #a #h #f #c #d #e #b #g #a #h #d #e #g #b #c #f #a #h #b #g #c #f #d #e ,the #a is here a repeater1 and the cycle #b1 #c1 #d1 #e1 #f1 #g1 #h4 ,every one of the #c groups will give #g lines4 ,here is my solution for #i persons on #bh occasions3-- #b #a #i #g #d #e #f #c #h #b #i #e #a #f #h #c #d #g #b #i #c #a #h #d #g #e #f #b #i #a #e #f #d #g #h #c ,there are here two repeaters1 #a and #b1 and the cycle is #c1 #d1 #e1 #f1 #g1 #h1 #i4 ,we thus get #d groups of #g lines each4 ,the case of #aj persons on #cf occasions is solved as follows3-- #a #aj #h #c #f #e #d #g #b #i #a #aj #f #e #b #i #g #d #c #h #a #aj #b #i #c #h #f #e #g #d #a #aj #g #d #h #c #b #i #e #f ,the repeater is #a1 and the cycle1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i1 #aj4 ,we here have #d groups of #i lines each4 ,my solution for #aa persons on #de occasions is as follows3-- #b #aa #i #d #g #f #e #a #h #c #aj #b #a #aa #g #f #c #aj #h #e #d #i #b #aa #aj #c #i #d #h #e #a #g #f #b #aa #e #h #a #c #aj #f #g #i #d #b #aa #a #aj #c #d #i #f #g #e #h ,there are two repeaters1 #a and #b1 and the cycle is1 #c1 #d1 #e1444 #aa4 ,we thus get #e groups of #i lines each4 ,the case of #ab persons on #ee occasions is solved thus3-- #a #b #c #ab #d #aa #e #aj #f #i #g #h #a #b #d #aa #f #i #h #g #aj #e #ab #c #a #b #e #aj #h #g #aa #d #c #ab #f #i #a #b #f #i #aj #e #c #ab #g #h #aa #d #a #b #g #h #ab #c #f #i #aa #d #e #aj ,here #a is a repeater1 and the cycle is #b1 #c1 #d1 #e1444 #ab4 ,we thus get #e groups of #aa lines each4 #bgd4--,,the ,,mouse-,,trap ,,puzzle4 ,if we interchange cards #f and #ac and begin our count at #ad1 we may take up all the twenty-one cards--that is1 make twenty-one catches--in the following order3 #f1 #h1 #ac1 #b1 #aj1 #a1 #aa1 #d1 #ad1 #c1 #e1 #g1 #ba1 #ab1 #ae1 #bj1 #i1 #af1 #ah1 #ag1 #ai4 ,we may also exchange #aj and #ad and start at #af1 or exchange #f and #h and start at #ai4 #bge4--,,the ,,sixteen ,,sheep4 ,the six diagrams on next page show solutions for the cases where we replace #b1 #c1 #d1 #e1 #f1 and #g hurdles4 ,the dark lines indicate the hurdles that have been replaced4 ,there are1 of course1 other ways of making the removals4 #bgf4--,,the ,,eight ,,villas4 ,there are several ways of solving the puzzle1 but there is very little difference between them4 ,the solver should1 however1 first of all bear in mind that in making his calculations he need only consider the four villas that stand at the corners1 because the intermediate villas can never vary when the corners are known4 ,one way is to place the numbers nought to #i one at a time in the top left-hand corner1 and then consider each case in turn4 ,now1 if we place #i in the corner as shown in the ,diagram ,a1 two of the corners cannot be occupied1 while the corner that is diagonally opposite may be filled by #j1 #a1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 or #i persons4 ,we thus see that there are #aj ,illustration3 ------ ------ ------ ,o ,,oho ,o ,,oho ,o ,o ,o ,,oho ,o ,h ,o ,,oho ,o ,,oho ,o ,o ,o ,,ohoho - -- ------ --- ,o,o ,o,o ,o,o ,o ,o ,o ,o ,o,o --- -- - ,o ,o ,o ,o ,o ,o ,,oho ,o ,o,o ,o ------- ------- ------- #b #c #d ------ ------ ------- ,o ,o ,,oho ,,oho ,o ,o ,o ,o ,o ,o ,o ,,oho ,o ,,ohoho ,o ,,ohoho ,o -- - ,o,o ,o,o ,o,o ,o,o ,o,,oho ,o ,o ,o ,o ,o ,,oho ,o ,o ,o ,o,o ,o ------- ------ ------ #e #f #g ,,the ,,sixteen ,,sheep solutions with a #i in the corner4 ,if1 however1 we substitute #h1 the two corners in the same row and column may contain #j1 #j1 or #a1 #a1 or #j1 #a1 or #a1 #j4 ,in the case of ,b1 ten different selections may be made for the fourth corner2 but in each of the cases ,c1 ,d1 and ,e1 only nine selections are possible1 because we cannot use the #i4 ,therefore with #h in the top left-hand corner there are #aj 7#c #i7 #cg different solutions4 ,if we then try #g in the corner1 the result will be #aj #bg #dj1 or #gg solutions4 ,with #f we get #aj #bg #dj #di #abf2 with #e1 #aj #bg #dj #di #ed #ahj2 with #d1 the same as with #e1 #ee #bce 2 with #c1 the same as with #d1 #eb #bhg2 with #b1 the same as with #c1 #de #ccb2 with #a1 the same as with #b1 #cd #cff1 and with nought in the top left-hand corner the number of solutions will be found to be #aj #bg #dj #di #ed #ee #eb #de #cd #ai #che4 ,as there is no other number to be placed in the top left-hand corner1 we have now only to add these totals together thus1 #aj #cg #gg #abf #ahj #bce #bhg #ccb #cff #che #b1jce4 ,we therefore find that the total number of ways in which tenants may occupy some or all of the eight villas so that there shall be always nine persons living along each side of the square is #b1jce4 ,of course1 this method must obviously cover all the reversals and reflections1 since each corner in turn is occupied by every number in all possible combinations with the other two corners that are in line with it4 ,illustration3 ,a ,b ,c ,d ,e --- --- --- --- --- #i #j #h #j #h #a #h #j #h #a --- --- --- --- --- 99 99 99 99 99 --- --- --- --- --- #j #j #a #a #j --- --- --- --- --- ,here is a general formula for solving the puzzle3 7n #cn #b77n #cn #c7_/f4 ,whatever may be the stipulated number of residents along each of the sides 7which number is represented by n71 the total number of different arrangements may be thus ascertained4 ,in our particular case the number of residents was nine4 ,therefore 7#ha #bg #b7 7#ha #bg #c7 and the product1 divided by #f1 gives #b1jce4 ,if the number of residents had been #j1 #a1 #b1 #c1 #d1 #e1 #f1 #g1 or #h1 the total arrangements would be #a1 #g1 #bf1 #gj1 #aee1 #cja1 #ecb1 #hgf1 or #a1cfe respectively4 #bgg4--;,,counter ,,crosses4 ,let us first deal with the ,greek ,cross4 ,there are just eighteen forms in which the numbers may be paired for the two arms4 ,here they are3-- #abigh #acifh #adieh #cdief #bdieg #bcifg #bcieh #acgfi #adgei #adifg #bdgeh #bcgfh #abehi #bcgei #acegi #cdefg #adgfh #bdefh #adefi #bcefi #adcgi #bcegh #adegh #becfh #aecfi #bdcfi #bcahi #bdcgh #aecgh #deafg #bdagi #beafi #cdafi #ceafh #cdagh #beagh ,of course1 the number in the middle is common to both arms4 ,the first pair is the one ,i gave as an example4 ,i will suppose that we have written out all these crosses1 always placing the first row of a pair in the upright and the second row in the horizontal arm4 ,now1 if we leave the central figure fixed1 there are #bd ways in which the numbers in the upright may be varied1 for the four counters may be changed in #a #b #c #d #bd ways4 ,and as the four in the horizontal may also be changed in #bd ways for every arrangement on the other arm1 we find that there are #bd #bd #egf variations for every form2 therefore1 as there are #ah forms1 we get #ah #egf #aj1cfh ways4 ,but this will include half the four reversals and half the four reflections that we barred1 so we must divide this by #d to obtain the correct answer to the ,greek ,cross1 which is thus #b1eib different ways4 ,the division is by #d and not by #h1 because we provided against half the reversals and reflections by always reserving one number for the upright and the other for the horizontal4 ,in the case of the ,latin ,cross1 it is obvious that we have to deal with the same #ah forms of pairing4 ,the total number of different ways in this case is the full number1 #ah #egf4 ,owing to the fact that the upper and lower arms are unequal in length1 permutations will repeat by reflection1 but not by reversal1 for we cannot reverse4 ,therefore this fact only entails division by #b4 ,but in every pair we may exchange the figures in the upright with those in the horizontal 7which we could not do in the case of the ,greek ,cross1 as the arms are there all alike72 consequently we must multiply by #b4 ,this multiplication by #b and division by #b cancel one another4 ,hence #aj1cfh is here the correct answer4 #bgh4--;,a ,,dormitory ,,puzzle4 ,illustration3 ,,mon4 ,,tues4 ,,wed4 --------- --------- --------- #a #b #a #a #c #a #a #d #a --------- --------- --------- #b #b #a #a #a #a --------- --------- --------- #a #bb #a #c #ai #c #d #af #d --------- --------- --------- ,,thurs4 ,,fri4 ,,sat4 --------- --------- --------- #a #e #a #b #f #b #d #d #d --------- --------- --------- #b #b #a #a #d #d --------- --------- --------- #d #ac #d #g #f #g #d #d #d --------- --------- --------- ,arrange the nuns from day to day as shown in the six diagrams4 ,the smallest possible number of nuns would be thirty-two1 and the arrangements on the last three days admit of variation4 #bgi4--,,the ,,barrels ,,of ,,balsam4 ,this is quite easy to solve for any number of barrels--if you know how4 ,this is the way to do it4 ,there are five barrels in each row ,multiply the numbers #a1 #b1 #c1 #d1 #e together2 and also multiply #f1 #g1 #h1 #i1 #aj together4 ,divide one result by the other1 and we get the number of different combinations or selections of ten things taken five at a time4 ,this is here #beb4 ,now1 if we divide this by #f 7#a more than the number in the row7 we get #db1 which is the correct answer to the puzzle1 for there are #db different ways of arranging the barrels4 ,try this method of solution in the case of six barrels1 three in each row1 and you will find the answer is #e ways4 ,if you check this by trial1 you will discover the five arrangements with #abc1 #abd1 #abe1 #acd1 #ace respectively in the top row1 and you will find no others4 ,the general solution to the problem is1 in fact1 this3 n ,c #bn ----- n #a where #bn equals the number of barrels4 ,the symbol ,c1 of course1 implies that we have to find how many combinations1 or selections1 we can make of #bn things1 taken n at a time4 #bhj4--;,,building ,,the ,,tetrahedron4 ,take your constructed pyramid and hold it so that one stick only lies on the table4 ,now1 four sticks must branch off from it in different directions--two at each end4 ,any one of the five sticks may be left out of this connection2 therefore the four may be selected in #e different ways4 ,but these four matches may be placed in #bd different orders4 ,and as any match may be joined at either of its ends1 they may further be varied 7after their situations are settled for any particular arrangement7 in #af different ways4 ,in every arrangement the sixth stick may be added in #b different ways4 ,now multiply these results together1 and we get #e #bd #af #b #c1hdj as the exact number of ways in which the pyramid may be constructed4 ,this method excludes all possibility of error4 ,a common cause of error is this4 ,if you calculate your combinations by working upwards from a basic triangle lying on the table1 you will get half the correct number of ways1 because you overlook the fact that an equal number of pyramids may be built on that triangle downwards1 so to speak1 through the table4 ,they are1 in fact1 reflections of the others1 and examples from the two sets of pyramids cannot be set up to resemble one another--except under fourth dimensional conditions6 #bha4--,,painting ,a ,,pyramid4 ,it will be convenient to imagine that we are painting our pyramids on the flat cardboard1 as in the diagrams1 before folding up4 ,now1 if we take any four colours 7say red1 blue1 green1 and yellow71 they may be applied in only #b distinctive ways1 as shown in ,figs1 #a and #b4 ,any other way will only result in one of these when the pyramids are folded up4 ,if we take any three colours1 they may be applied in the #c ways shown in ,figs4 #c1 #d1 and #e4 ,if we take any two colours1 they may be applied in the #c ways shown in ,figs4 #f1 #g1 and #h4 ,if we take any single colour1 it may obviously be applied in only #a way4 ,but four colours may be selected in #ce ways out of seven2 three in #ce ways2 two in #ba ways2 and one colour in #g ways4 ,therefore #ce applied in #b ways #gj2 #ce in #c ways #aje2 #ba in #c ways #fc2 and #g in #a way #g4 ,consequently the pyramid may be painted in #bde different ways 7#gj #aje #fc #g71 using the seven colours of the solar spectrum in accordance with the conditions of the puzzle4 ,illustration3 #a #b --------------- --------------- ,r _/ ,b _/ ,b _/ ,r _/ _/ _/ _/ _/ _/ ,g _/ _/ ,g _/ -------_/ -------_/ _/ _/ ,y _/ ,y _/ _/ _/ ' ' #c #d #e --------------- --------------- --------------- ,r _/ ,r _/ ,r _/ ,g _/ ,y _/ ,r _/ _/ _/ _/ _/ _/ _/ _/ ,g _/ _/ ,g _/ _/ ,g _/ -------_/ -------_/ -------_/ _/ _/ _/ ,y _/ ,y _/ ,y _/ _/ _/ _/ ' ' ' #f #g #h --------------- --------------- --------------- ,g _/ ,y _/ ,y _/ ,y _/ ,g _/ ,g _/ _/ _/ _/ _/ _/ _/ _/ ,g _/ _/ ,g _/ _/ ,g _/ -------_/ -------_/ -------_/ _/ _/ _/ ,y _/ ,y _/ ,y _/ _/ _/ _/ ' ' ' #bhb4--,,the ,,antiquary',s ,,chain4 ,illustration ,,the number of ways in which nine things may be arranged in a row without any restrictions is #a #b #c #d #e #f #g #h #i #cfb1hhj4 ,but we are told that the two circular rings must never be together2 therefore we must deduct the number of times that this would occur4 ,the number is #a #b #c #d #e #f #g #h #dj1cbj #b #hj1fdj1 because if we consider the two circular links to be inseparably joined together they become as one link1 and eight links are capable of #dj1cbj arrangements2 but as these two links may always be put on in the orders ,,ab or ,,ba1 we have to double this number1 it being a question of arrangement and not of design4 ,the deduction required reduces our total to #bhb1bdj4 ,then one of our links is of a peculiar form1 like an #h4 ,we have therefore the option of joining on either one end or the other on every occasion1 so we must double the last result4 ,this brings up our total to #efd1dhj4 ,we now come to the point to which ,i directed the reader's attention--that every link may be put on in one of two ways4 ,if we join the first finger and thumb of our left hand horizontally1 and then link the first finger and thumb of the right hand1 we see that the right thumb may be either above or below4 ,but in the case of our chain we must remember that although that #h-shaped link has two independent ends it is like every other link in having only two sides--that is1 you cannot turn over one end without turning the other at the same time4 ,we will1 for convenience1 assume that each link has a black side and a side painted white4 ,now1 if it were stipulated that 7with the chain lying on the table1 and every successive link falling over its predecessor in the same way1 as in the diagram7 only the white sides should be uppermost as in ,a1 then the answer would be #efd1dhj1 as above--ignoring for the present all reversals of the completed chain4 ,if1 however1 the first link were allowed to be placed either side up1 then we could have either ,a or ,b1 and the answer would be #b #efd1dhj #a1abh1ifj2 if two links might be placed either way up1 the answer would be #d #efd1dhj2 if three links1 then #h #efd1dhj1 and so on4 ,since1 therefore1 every link may be placed either side up1 the number will be #efd1dhj multiplied by #bi1 or by #eab4 ,this raises our total to #bhi1jac1gfj4 ,but there is still one more point to be considered4 ,we have not yet allowed for the fact that with any given arrangement three of the other arrangements may be obtained by simply turning the chain over through its entire length and by reversing the ends4 ,thus ,c is really the same as ,a1 and if we turn this page upside down1 then ,a and ,c give two other arrangements that are still really identical4 ,thus to get the correct answer to the puzzle we must divide our last total by #d1 when we find that there are just #gb1bec1ddj different ways in which the smith might have put those links together4 ,in other words1 if the nine links had originally formed a piece of chain1 and it was known that the two circular links were separated1 then it would be #gb1bec1dci chances to #a that the smith would not have put the links together again precisely as they were arranged before6 #bhc4--,,the ,,fifteen ,,dominoes4 ,the reader may have noticed that at each end of the line ,i give is a four1 so that1 if we like1 we can form a ring instead of a line4 ,it can easily be proved that this must always be so4 ,every line arrangement will make a circular arrangement if we like to join the ends4 ,now1 curious as it may at first appear1 the following diagram exactly represents the conditions when we leave the doubles out of the question and devote our attention to forming circular arrangements4 ,each number1 or half domino1 is in line with every other number1 so that if we start at any one of the five numbers and go over all the lines of the pentagon once and once only we shall come back to the starting place1 and the order of our route will give us one of the circular arrangements for the ten dominoes4 ,take your pencil and follow out the following route1 starting at the #d3 #dacjdbajbcd4 ,you have been over all the lines once only1 and by repeating all these figures in this way1 #da--ac--cj--jd--db--ba--aj--jb--bc--cd1 you get an arrangement of the dominoes 7without the doubles7 which will be perfectly clear4 ,take other routes and you will get other arrangements4 ,if1 therefore1 we can ascertain just how many of these circular routes are obtainable from the pentagon1 then the rest is very easy4 ,well1 the number of different circular routes over the pentagon is #bfd4 ,how ,i arrive at these figures ,i will not at present explain1 because it would take a lot of space4 ,the dominoes may1 therefore1 be arranged in a circle in just #bfd different ways1 leaving out the doubles4 ,now1 in any one of these circles the five doubles may be inserted in #be #cb different ways4 ,therefore when we include the doubles there are #bfd #cb #h1ddh different circular arrangements4 ,but each of those circles may be broken 7so as to form our straight line7 in any one of #ae different places4 ,consequently1 #h1ddh #ae gives #abf1gbj different ways as the correct answer to the puzzle4 ,illustration3 ----- _/ _/ ----- _/ 4 4 ----- 4 4 ----- 4 4 o o o -4--------4--- 4 4 4 o o ----- 4 4 44 ----- 4 4 4 4 _/ ----- 44 ----- o 4 4 o --------- o o 4 4 o ----- ----- ,i purposely refrained from asking the reader to discover in just how many different ways the full set of twenty-eight dominoes may be arranged in a straight line in accordance with the ordinary rules of the game1 left to right and right to left of any arrangement counting as different ways4 ,it is an exceedingly difficult problem1 but the correct answer is #g1iei1bbi1ica1ebj ways4 ,the method of solving is very complex4 #bhd4--,,the ,,cross ,,target4 ,illustration3 -- -- 7,,cd77 7 -- -- 7,,ae77,a 7 -- -- -- -- -- -- 7,,ce77,e 77,a 77,,ab77,c 77,d 7 -- -- -- -- -- -- 7,d 77 77,b 77,e 77,,eb77 7 -- -- -- -- -- -- 7,c 77,b 7 -- -- 7 77,,ed7 -- -- ,twenty-one different squares may be selected4 ,of these nine will be of the size shown by the four ,a's in the diagram1 four of the size shown by the ,b's1 four of the size shown by the ,c's1 two of the size shown by the ,d's1 and two of the size indicated by the upper single ,a1 the upper single ,e1 the lower single ,c1 and the ,,eb4 ,it is an interesting fact that you cannot form any one of these twenty-one squares without using at least one of the six circles marked ,e4 #bhe4--,,the ,,four ,,postage ,,stamps4 ,referring to the original diagram1 the four stamps may be given in the shape #a1 #b1 #c1 #d1 in three ways2 in the shape #a1 #b1 #e1 #f1 in six ways2 in the shape #a1 #b1 #c1 #e1 or #a1 #b1 #c1 #g1 or #a1 #e1 #f1 #g1 or #c1 #e1 #f1 #g1 in twenty-eight ways2 in shape #a1 #b1 #c1 #f1 or #b1 #e1 #f1 #g1 in fourteen ways2 in shape #a1 #b1 #f1 #g1 or #b1 #c1 #e1 #f1 or #a1 #e1 #f1 #aj1 or #b1 #e1 #f1 #i1 in fourteen ways4 ,thus there are sixty-five ways in all4 #bhf4--,,painting ,,the ,,die4 ,the #a can be marked on any one of six different sides4 ,for every side occupied by #a we have a selection of four sides for the #b4 ,for every situation of the #b we have two places for the #c4 7,the #f1 #e1 and #d need not be considered1 as their positions are determined by the #a1 #b1 and #c47 ,therefore #f1 #d1 and #b multiplied together make #dh different ways--the correct answer4 #bhg4--;,,an ,,acrostic ,,puzzle4 ,there are twenty-six letters in the alphabet1 giving #cbe different pairs4 ,every one of these pairs may be reversed1 making #fej ways4 ,but every initial letter may be repeated as the final1 producing #bf other ways4 ,the total is therefore #fgf different pairs4 ,in other words1 the answer is the square of the number of letters in the alphabet4 #bhh4--;,,chequered ,,board ,,divisions4 ,there are #bee different ways of cutting the board into two pieces of exactly the same size and shape4 ,every way must involve one of the five cuts shown in ,diagrams ,a1 ,b1 ,c1 ,d1 and ,e4 ,to avoid repetitions by reversal and reflection1 we need only consider cuts that enter at the points a1 b1 and c4 ,but the exit must always be at a point in a straight line from the entry through the centre4 ,this is the most important condition to remember4 ,in case ,b you cannot enter at a1 or you will get the cut provided for in ,e4 ,similarly in ,c or ,d1 you must not enter the key-line in the same direction as itself1 or you will get ,a or ,b4 ,if you are working on ,a or ,c and entering at a1 you must consider joins at one end only of the key-line1 or you will get repetitions4 ,in other cases you must consider joins at both ends of the key2 but after leaving a in case ,d1 turn always either to right or left--use one direction only4 ,figs4 #a and #b are examples under ,a2 #c and #d are examples under ,b2 #e and #f come under ,c2 ,illustration and #g is a pretty example of ,d4 ,of course1 ,e is a peculiar type1 and obviously admits of only one way of cutting1 for you clearly cannot enter at b or c4 ,here is a table of the results3-- a b c ,ways4 ,a #h #ag #ba #df ,b #j #ag #ba #ch ,c #ae #ca #ci #he ,d #ag #bi #ci #he ,e #a #j #j #a -- -- -- --- #da #id #abj #bee ,i have not attempted the task of enumerating the ways of dividing a board #h #h--that is1 an ordinary chessboard4 ,whatever the method adopted1 the solution would entail considerable labour4 #bhi4--,,lions ,,and ,,crowns4 ,illustration ,here is the solution4 ,it will be seen that each of the four pieces 7after making the cuts along the thick lines7 is of exactly the same size and shape1 and that each piece contains a lion and a crown4 ,two of the pieces are shaded so as to make the solution quite clear to the eye4 #bij4--;,,boards ,,with ,,an ,,odd ,,number ,,of ,,squares4 ,there are fifteen different ways of cutting the #e #e board 7with the central square removed7 into two pieces of the same size and shape4 ,limitations of space will not allow me to give diagrams of all these1 but ,i will enable the reader to draw them all out for himself without the slightest difficulty4 ,at whatever point on the edge your cut enters1 it must always end at a point on the edge1 exactly opposite in a line through the centre of the square4 ,thus1 if you enter at point #a 7see ,fig4 #a7 at the top1 you must leave at point #a at the bottom4 ,now1 #a and #b are the only two really different points of entry2 if we use any others they will simply produce similar solutions4 ,the directions of the cuts in the following fifteen ,illustration3 ,fig4 #a4 ,fig4 #b4 solutions are indicated by the numbers on the diagram4 ,the duplication of the numbers can lead to no confusion1 since every successive number is contiguous to the previous one4 ,but whichever direction you take from the top downwards you must repeat from the bottom upwards1 one direction being an exact reflection of the other4 #a1 #d1 #h4 #a1 #d1 #c1 #g1 #h4 #a1 #d1 #c1 #g1 #aj1 #i4 #a1 #d1 #c1 #g1 #aj1 #f1 #e1 #i4 #a1 #d1 #e1 #i4 #a1 #d1 #e1 #f1 #aj1 #i4 #a1 #d1 #e1 #f1 #aj1 #g1 #h4 #b1 #c1 #d1 #h4 #b1 #c1 #d1 #e1 #i4 #b1 #c1 #d1 #e1 #f1 #aj1 #i4 #b1 #c1 #d1 #e1 #f1 #aj1 #g1 #h4 #b1 #c1 #g1 #h4 #b1 #c1 #g1 #aj1 #i4 #b1 #c1 #g1 #aj1 #f1 #e1 #i4 #b1 #c1 #g1 #aj1 #f1 #e1 #d1 #h4 ,it will be seen that the fourth direction 7#a1 #d1 #c1 #g1 #aj1 #f1 #e1 #i7 produces the solution shown in ,fig4 #b4 ,the thirteenth produces the solution given in propounding the puzzle1 where the cut entered at the side instead of at the top4 ,the pieces1 however1 will be of the same shape if turned over1 which1 as it was stated in the conditions1 would not constitute a different solution4 #bia4--,,the ,,grand ,,lama',s ,,problem4 ,the method of dividing the chessboard so that each of the four parts shall be of exactly the same size and shape1 and contain one of the gems1 is shown in the diagram4 ,the method of shading the squares is adopted to make the shape of the pieces clear to the eye4 ,two of the pieces are shaded and two left white4 ,the reader may find it interesting to compare this puzzle with that of the ,weaver 7,no4 #ad1 ,canterbury ,puzzles74 ,illustration3 ,,the ,,grand ,,lama',s ,,problem4 3o3 3 3 3 3 3 3 3 ,i444,i444 333 o 33333333333333333333333 ,i444,i444,i444444,i 333 3o3 3 3 3 333 ,i444,i444,i444,i444,i444,i444,i 333 333 o 3333333 333 ,i444,i444,i444,i444,i444,i444,i 333 3333333 333 333 ,i444,i444444444,i444,i444,i 333 3 3 3 333 333 ,i444,i444,i444,i444,i 33333333333333333333333 333 444,i444,i 3 3 3 3 3 3 333 #bib4--,,the ,,abbot',s ,,window4 ,,the man who was learned in strange mysteries pointed out to ,father ,john that the orders of the ,lord ,abbot of ,st4 ,edmondsbury might be easily carried out by blocking up twelve of the lights in the window as shown by the dark squares in the following sketch3-- ,illustration3 3 3 3 3 3 3 3 ,i444444444444444444,i ,,iiiii 3 3 3 ,,iiiii ,i444444444444,i 3 ,,iiiii 3 ,,iiiii 3 ,i444444444444,i 3 3 ,,iiiiiiiii 3 3 ,i444444444444444444,i 3 3 ,,iiiiiiiii 3 3 ,i444444444444,i 3 ,,iiiii 3 ,,iiiii 3 ,i444444444444,i ,,iiiii 3 3 3 ,,iiiii ,i444444444444444444,i 3 3 3 3 3 3 3 ,father ,john held that the four corners should also be darkened1 but the sage explained that it was desired to obstruct no more light than was absolutely necessary1 and he said1 anticipating ,lord ,dundreary1 ,a single pane can no more be in a line with itself than one bird can go into a corner and flock in solitude4 ,the ,abbot's condition was that no diagonal lines should contain an odd number of lights4 ,now1 when the holy man saw what had been done he was well pleased1 and said1 ,truly1 ,father ,john1 thou art a man of deep wisdom1 in that thou hast done that which seemed impossible1 and yet withal adorned our window with a device of the cross of ,st4 ,andrew1 whose name ,i received from my godfathers and godmothers4 ,thereafter he slept well and arose refreshed4 ,the window might be seen intact to-day in the monastery of ,st4 ,edmondsbury1 if it existed1 which1 alas6 the window does not4 #bic4--,,the ,,chinese ,,chessboard4 ,i,i 3333 #b 3333 #c 333 #e 3#f3 ,i444444444,i444,i444,i 3333 #a 333 3333 #d 333 #g ,i444444,i,i444,i,i 3333 333 3333 #i 333 ,i,i444,i,i444,i444,i 3333 #aa3333 33333 #aj333 #h ,i,i,i,i444,i444,i 33333 #ab3333 #ac3333 333 ,i444,i444,i 3333 #ad333 333 #af3333 #ag ,i444444,i,i444444444,i 3333 33333 #ae333 3333 ,i,i 3333 33333 #ah33333 33333 ,i,i,i 3333 3333 333 3333 ,i444,i444,i,i444,i444,i 333 3333 3333 333 ,i444,i,i,i,i444,i 3333 3333 33333 333 ,i,i,i,i,i,i 333 3333 3333 3333 ,i444,i444,i444444,i444444,i 3333 3333 333 3333 ,i444,i444,i444,i444,i 333 3333 3333 333 ,i,i444,i444,i,i 3333 3333 33333 333 ,i444444,i444444,i 3333 3333 333 33333 ,eighteen is the maximum number of pieces4 ,i give two solutions4 ,the numbered diagram is so cut that the eighteenth piece has the largest area--eight squares--that is possible under the conditions4 ,the second diagram was prepared under the added condition that no piece should contain more than five squares4 ,no4 #gd in ,the ,canterbury ,puzzles shows how to cut the board into twelve pieces1 all different1 each containing five squares1 with one square piece of four squares4 #bid4--,,the ,,chessboard ,,sentence4 ,i,i,i,i,i 333 333 3333 3333 ,i,i444,i,i444,i444,i444,i 333 33333 333 33333 444444,i444,i444444,i 333 333 3333 3333 444444,i,i,i,i 3333 33333 333 33333 ,i,i,i444,i,i444 33333 333 333 333 444444444444444,i444 333 333 333 3333 444444444444,i444444 333 333 33333 333 ,i444,i444,i444 3333 3333 33333 3333 ,i ,the pieces may be fitted together1 as shown in the illustration1 to form a perfect chessboard4 #bie4--,,the ,,eight ,,rooks4 ,obviously there must be a rook in every row and every column4 ,starting with the top row1 it is clear that we may put our first rook on any one of eight different squares4 ,wherever it is placed1 we have the option of seven squares for the second rook in the second row4 ,then we have six squares from which to select the third row1 five in the fourth1 and so on4 ,therefore the number of our different ways must be #h #g #f #e #d #c #b #a #dj1cbj 7that is #h671 which is the correct answer4 ,how many ways there are if mere reversals and reflections are not counted as different has not yet been determined2 it is a difficult problem4 ,but this point1 on a smaller square1 is considered in the next puzzle4 #bif4--,,the ,,four ,,lions4 ,there are only seven different ways under the conditions4 ,they are as follows3 #a #b #c #d1 #a #b #d #c1 #a #c #b #d1 #a #c #d #b1 #a #d #c #b1 #b #a #d #c1 #b #d #a #c4 ,taking the last example1 this notation means that we place a lion in the second square of first row1 fourth square of second row1 first square of third row1 and third square of fourth row4 ,the first example is1 of course1 the one we gave when setting the puzzle4 #big4--;,,bishops--,,unguarded4 444444444444444444444444 3 33333 33333 33333 33333 444444444444444444444444 33333 33333 33333 33333 3 444444444444444444444444 3 33333 33333 33333 33333 444444444444444444444444 33,b33 ,b 33,b33 ,b 33,b33 ,b 33,b33 ,b 3 444444444444444444444444 3 33333 33333 33333 33333 444444444444444444444444 33333 33333 33333 33333 3 444444444444444444444444 33333 33333 33333 33333 3 444444444444444444444444 3 33333 33333 33333 33333 444444444444444444444444 ,this cannot be done with fewer bishops than eight1 and the simplest solution is to place the bishops in line along the fourth or fifth row of the board 7see diagram74 ,but it will be noticed that no bishop is here guarded by another1 so we consider that point in the next puzzle4 #bih4--;,,bishops--,,guarded4 44444444444444444444444444 3 33333 33333 33333 33333 444444444444444444444444 33333 33333 33333 33333 3 444444444444444444444444 3 33333 33333 33333 33333 4444444444444444444444444 33333 ,b 33,b33 ,b 33333 ,b 33,b33 3 44444444444444444444444444 3 33,b33 ,b 33,b33 33,b33 ,b 33333 444444444444444444444444 33333 33333 33333 33333 3 444444444444444444444444 3 33333 33333 33333 33333 4444444444444444444444444 33333 33333 33333 33333 3 4444444444444444444444444 ,this puzzle is quite easy if you first of all give it a little thought4 ,you need only consider squares of one colour1 for whatever can be done in the case of the white squares can always be repeated on the black1 and they are here quite independent of one another4 ,this equality1 of course1 is in consequence of the fact that the number of squares on an ordinary chessboard1 sixty-four1 is an even number4 ,if a square chequered board has an odd number of squares1 then there will always be one more square of one colour than of the other4 ,ten bishops are necessary in order that every square shall be attacked and every bishop guarded by another bishop4 ,i give one way of arranging them in the diagram4 ,it will be noticed that the two central bishops in the group of six on the left-hand side of the board serve no purpose1 except to protect those bishops that are on adjoining squares4 ,another solution would therefore be obtained by simply raising the upper one of these one square and placing the other a square lower down4 #bii4--;,,bishops ,,in ,,convocation4 ,the fourteen bishops may be placed in #bef different ways4 ,but every bishop must always be placed on one of the sides of the board--that is1 somewhere on a row or file on the extreme edge4 ,the puzzle1 therefore1 consists in counting the number of different ways that we can arrange the fourteen round the edge of the board without attack4 ,this is not a difficult matter4 ,on a chessboard of n squares #bn - #b bishops 7the maximum number7 may always be placed in #bn ways without attacking4 ,on an ordinary chessboard n would be #h2 therefore #ad bishops may be placed in #bef different ways4 ,it is rather curious that the general result should come out in so simple a form4 ,illustration #cjj4--,,the ,,eight ,,queens4 ,illustration ,the solution to this puzzle is shown in the diagram4 ,it will be found that no queen attacks another1 and also that no three queens are in a straight line in any oblique direction4 ,this is the only arrangement out of the twelve fundamentally different ways of placing eight queens without attack that fulfils the last condition4 #cja4--,,the ,,eight ,,stars4 ,the solution of this puzzle is shown in the first diagram4 ,it is the only possible solution within the conditions stated4 ,but if one of the eight stars had not already been placed as shown1 there would then have been eight ways of arranging the stars according to this scheme1 if we count reversals and reflections as different4 ,if you turn this page round so that each side is in turn at the bottom1 you will get the four reversals2 and if you reflect each of these in a mirror1 you will get the four reflections4 ,these are1 therefore1 merely eight aspects of one fundamental solution4 ,but without that first star being so placed1 there is another fundamental solution1 as shown in the second diagram4 ,but this arrangement being in a way symmetrical1 only produces four different aspects by reversal and reflection4 ,illustration #cjb4--;,a ,,problem ,,in ,,mosaics4 ,illustration ,the diagram shows how the tiles may be rearranged4 ,as before1 one yellow and one purple tile are dispensed with4 ,i will here point out that in the previous arrangement the yellow and purple tiles in the seventh row might have changed places1 but no other arrangement was possible4 #cjc4--,,under ,,the ,,veil4 ,some schemes give more diagonal readings of four letters than others1 and we are at first tempted to favour these2 but this is a false scent1 because what you appear to gain in this direction you lose in others4 ,of course it immediately occurs to the solver that every ,,live or ,,evil is worth twice as much as any other word1 since it reads both ways and always counts as #b4 ,this is an important consideration1 though sometimes those arrangements that contain most readings of these two words are fruitless in other words1 and we lose in the general count4 ,illustration3 ,i ,v ,e ,l ,e ,v ,l ,i ,l ,i ,v ,e ,i ,v ,e ,l ,e ,l ,v ,i ,l ,i ,i ,e ,v _/,v ,e ,l ,i ,i ,v ,e ,l ,the above diagram is in accordance with the conditions requiring no letter to be in line with another similar letter1 and it gives twenty readings of the five words--six horizontally1 six vertically1 four in the diagonals indicated by the arrows on the left1 and four in the diagonals indicated by the arrows on the right4 ,this is the maximum4 ,four sets of eight letters may be placed on the board of sixty-four squares in as many as #fjd different ways1 without any letter ever being in line with a similar one4 ,this does not count reversals and reflections as different1 and it does not take into consideration the actual permutations of the letters among themselves2 that is1 for example1 making the ,l's change places with the ,e's4 ,now it is a singular fact that not only do the twenty word-readings that ,i have given prove to be the real maximum1 but there is actually only that one arrangement from which this maximum may be obtained4 ,but if you make the ,v's change places with the ,i's1 and the ,l's with the ,e's1 in the solution given1 you still get twenty readings--the same number as before in every direction4 ,therefore there are two ways of getting the maximum from the same arrangement4 ,the minimum number of readings is zero--that is1 the letters can be so arranged that no word can be read in any of the directions4 #cjd4--;,,bachet',s ,,square4 ,illustration3 #a ,illustration3 #b ,illustration3 #c ,illustration3 #d ,let us use the letters ,a1 ,k1 ,q1 ,j1 to denote ace1 king1 queen1 jack2 and ,d1 ,s1 ,h1 ,c1 to denote diamonds1 spades1 hearts1 clubs4 ,in ,diagrams #a and #b we have the two available ways of arranging either group of letters so that no two similar letters shall be in line--though a quarter-turn of #a will give us the arrangement in #b4 ,if we superimpose or combine these two squares1 we get the arrangement of ,diagram #c1 which is one solution4 ,but in each square we may put the letters in the top line in twenty-four different ways without altering the scheme of arrangement4 ,thus1 in ,diagram #d the ,s's are similarly placed to the ,d's in #b1 the ,h's to the ,s's1 the ,c's to the ,h's1 and the ,d's to the ,c's4 ,it clearly follows that there must be #bdbd #egf ways of combining the two primitive arrangements4 ,but the error that ,labosne fell into was that of assuming that the ,a1 ,k1 ,q1 ,j must be arranged in the form #a1 and the ,d1 ,s1 ,h1 ,c in the form #b4 ,he thus included reflections and half-turns1 but not quarter-turns4 ,they may obviously be interchanged4 ,so that the correct answer is #b #egf #a1aeb1 counting reflections and reversals as different4 ,put in another manner1 the pairs in the top row may be written in #af #i #d #a #egf different ways1 and the square then completed in #b ways1 making #a1aeb ways in all4 #cje4--,,the ,,thirty-,,six ,,letter ,,blocks4 ,i pointed out that it was impossible to get all the letters into the box under the conditions1 but the puzzle was to place as many as possible4 ,this requires a little judgment and careful investigation1 or we are liable to jump to the hasty conclusion that the proper way to solve the puzzle must be first to place all six of one letter1 then all six of another letter1 and so on4 ,as there is only one scheme 7with its reversals7 for placing six similar letters so that no two shall be in a line in any direction1 the reader will find that after he has placed four different kinds of letters1 six times each1 every place is occupied except those twelve that form the two long diagonals4 ,he is1 therefore1 unable to place more than two each of his last two letters1 and there are eight blanks left4 ,i give such an arrangement in ,diagram #a4 ,illustration3 #a ,illustration3 #b ,the secret1 however1 consists in not trying thus to place all six of each letter4 ,it will be found that if we content ourselves with placing only five of each letter1 this number 7thirty in all7 may be got into the box1 and there will be only six blanks4 ,but the correct solution is to place six of each of two letters and five of each of the remaining four4 ,an examination of ,diagram #b will show that there are six each of ,c and ,d1 and five each of ,a1 ,b1 ,e1 and ,f4 ,there are1 therefore1 only four blanks left1 and no letter is in line with a similar letter in any direction4 #cjf4--,,the ,,crowded ,,chessboard4 ,illustration ,here is the solution4 ,only #h queens or #h rooks can be placed on the board without attack1 while the greatest number of bishops is #ad1 and of knights #cb4 ,but as all these knights must be placed on squares of the same colour1 while the queens occupy four of each colour and the bishops #g of each colour1 it follows that only #ba knights can be placed on the same colour in this puzzle4 ,more than #ba knights can be placed alone on the board if we use both colours1 but ,i have not succeeded in placing more than #ba on the crowded chessboard4 ,i believe the above solution contains the maximum number of pieces1 but possibly some ingenious reader may succeed in getting in another knight4 #cjg4--,,the ,,coloured ,,counters4 ,the counters may be arranged in this order3-- ,r#a1 ,b#b1 ,y#c1 ,o#d1 ,,gs4 ,y#d1 ,o#e1 ,g#a1 ,r#b1 ,b#c4 ,g#b1 ,r#c1 ,b#d1 ,y#e1 ,o#a4 ,b#e1 ,y#a1 ,o#b1 ,g#c1 ,r#d4 ,o#c1 ,g#d1 ,r#e1 ,b#a1 ,y#b4 #cjh4--,,the ,,gentle ,,art ,,of ,,stamp-,,licking4 ,the following arrangement shows how sixteen stamps may be stuck on the card1 under the conditions1 of a total value of fifty pence1 or #ds4 #b;d43-- ,illustration ,if1 after placing the four #e;d4 stamps1 the reader is tempted to place four #d;d4 stamps also1 he can afterwards only place two of each of the three other denominations1 thus losing two spaces and counting no more than forty-eight pence1 or #ds4 ,this is the pitfall that was hinted at4 7,compare with ,no4 #dc1 ,canterbury ,puzzles47 #cji4--,,the ,,forty-,,nine ,,counters4 ,the counters may be arranged in this order3-- ,a#a1 ,b#b1 ,c#c1 ,d#d1 ,e#e1 ,f#f1 ,g#g4 ,f#d1 ,g#e1 ,a#f1 ,b#g1 ,c#a1 ,d#b1 ,e#c4 ,d#g1 ,e#a1 ,f#b1 ,g#c1 ,a#d1 ,b#e1 ,c#f4 ,b#c1 ,c#d1 ,d#e1 ,e#f1 ,f#g1 ,g#a1 ,a#b4 ,g#f1 ,a#g1 ,b#a1 ,c#b1 ,d#c1 ,e#d1 ,f#e4 ,e#b1 ,f#c1 ,g#d1 ,a#e1 ,b#f1 ,c#g1 ,d#a4 ,c#e1 ,d#f1 ,e#g1 ,f#a1 ,g#b1 ,a#c1 ,b#d4 #caj4--,,the ,,three ,,sheep4 ,the number of different ways in which the three sheep may be placed so that every pen shall always be either occupied or in line with at least one sheep is forty-seven4 ,the following table1 if used with the key in ,diagram #a1 will enable the reader to place them in all these ways3-- ---------------------------------------- --------- ,no4 of ,two ,sheep4 ,third ,sheep4 ,ways4 ---------------------------------------- --------- ,a and ,b ,c1 ,e1 ,g1 ,k1 ,l1 ,n1 or ,p #g ,a and ,c ,i1 ,j1 ,k1 or ,o #d ,a and ,d ,m1 ,n1 or ,j #c ,a and ,f ,j1 ,k1 ,l1 or ,p #d ,a and ,g ,h1 ,j1 ,k1 ,n1 ,o1 or ,p #f ,a and ,h ,k1 ,l1 ,n1 or ,o #d ,a and ,o ,k or ,l #b ,b and ,c ,n #a ,b and ,e ,f1 ,h1 ,k1 or ,l #d ,b and ,f ,g1 ,j1 ,n1 or ,o #d ,b and ,g ,k1 ,l1 or ,n #c ,b and ,h ,j or ,n #b ,b and ,j ,k or ,l #b ,f and ,g ,j #a ---- #dg ---------------------------------------- --------- ,this1 of course1 means that if you place sheep in the pens marked ,a and ,b1 then there are seven different pens in which you may place the third sheep1 giving seven different solutions4 ,it was understood that reversals and reflections do not count as different4 ,if one pen at least is to be not in line with a sheep1 there would be thirty solutions to that problem4 ,if we counted all the reversals and reflections of these #dg and #cj cases respectively as different1 their total would be #efj1 which is the number of different ways in which the sheep may be placed in three pens without any conditions4 ,i will remark that there are three ways in which two sheep may be placed so that every pen is occupied or in line1 as in ,diagrams #b1 #c1 and #d1 but in every case each sheep is in line with its companion4 ,there are only two ways in which three sheep may be so placed that every pen shall be occupied or in line1 but no sheep in line with another4 ,these ,i show in ,diagrams #e and #f4 ,finally1 there is only one way in which three sheep may be placed so that at least one pen shall not be in line with a sheep and yet no sheep in line with another4 ,place the sheep in ,c1 ,e1 ,l4 ,this is practically all there is to be said on this pleasant pastoral subject4 ,illustration #caa4--,,the ,,five ,,dogs ,,puzzle4 ,the diagrams show four fundamentally different solutions4 ,in the case of ,a we can reverse the order1 so that the single dog is in the bottom row and the other four shifted up two squares4 ,also we may use the next column to the right and both of the two central horizontal rows4 ,thus ,a gives #h solutions4 ,then ,b may be reversed and placed in either diagonal1 giving #d solutions4 ,similarly ,c will give #d solutions4 ,the line in ,d being symmetrical1 its reversal will not be different1 but it may be disposed in #d different directions4 ,we thus have in all #bj different solutions4 ,illustration #cab4--,,the ,,five ,,crescents ,,of ,,byzantium4 ,illustration ,if that ancient architect had arranged his five crescent tiles in the manner shown in the following diagram1 every tile would have been watched over by1 or in a line with1 at least one crescent1 and space would have been reserved for a perfectly square carpet equal in area to exactly half of the pavement4 ,it is a very curious fact that1 although there are two or three solutions allowing a carpet to be laid down within the conditions so as to cover an area of nearly twenty-nine of the tiles1 this is the only possible solution giving exactly half the area of the pavement1 which is the largest space obtainable4 #cac4--,,queens ,,and ,,bishop ,,puzzle4 ,illustration3 ,,fig4 #a4 ,illustration3 ,,fig4 #b4 ,the bishop is on the square originally occupied by the rook1 and the four queens are so placed that every square is either occupied or attacked by a piece4 7,fig4 #a47 ,i pointed out in #ahii that if four queens are placed as shown in the diagram 7,fig4 #b71 then the fifth queen may be placed on any one of the twelve squares marked a1 b1 c1 d1 and e2 or a rook on the two squares1 c2 or a bishop on the eight squares1 a1 b1 and e2 or a pawn on the square b2 or a king on the four squares1 b1 c1 and e4 ,the only known arrangement for four queens and a knight is that given by ,mr4 ,j4 ,wallis in ,the ,strand ,magazine for ,august #aijh1 here reproduced4 7,fig4 #c47 ,illustration3 ,,fig4 #c4 ,i have recorded a large number of solutions with four queens and a rook1 or bishop1 but the only arrangement1 ,i believe1 with three queens and two rooks in which all the pieces are guarded is that of which ,i give an illustration 7,fig4 #d71 first published by ,dr4 ,c4 ,planck4 ,but ,i have since found the accompanying solution with three queens1 a rook1 and a bishop1 though the pieces do not protect one another4 7,fig4 #e47 ,illustration3 ,,fig4 #d4 ,illustration3 ,,fig4 #e4 #cad4--,,the ,,southern ,,cross4 ,my readers have been so familiarized with the fact that it requires at least five planets to attack every one of a square arrangement of sixty-four stars that many of them have1 perhaps1 got to believe that a larger square arrangement of stars must need an increase of planets4 ,it was to correct this possible error of reasoning1 and so warn readers against another of those numerous little pitfalls in the world of puzzledom1 that ,i devised this new stellar problem4 ,let me then state at once that1 in the case of a square arrangement of eighty one stars1 there are several ways of placing five planets so that every star shall be in line with at least one planet vertically1 horizontally1 or diagonally4 ,here is the solution to the ,southern ,cross3 -- ,it will be remembered that ,i said that the five planets in their new positions will1 of course1 obscure five other stars in place of those at present covered4 ,this was to exclude an easier solution in which only four planets need be moved4 #cae4--,,the ,,hat-,,peg ,,puzzle4 ,the moves will be made quite clear by a reference to the diagrams1 which show the position on the board after each of the four moves4 ,the darts indicate the successive removals that have been made4 ,it will be seen that at every stage all the squares are either attacked or occupied1 and that after the fourth move no queen attacks any other4 ,in the case of the last move the queen in the top row might also have been moved one square farther to the left4 ,this is1 ,i believe1 the only solution to the puzzle4 ,illustration3 #a ,illustration3 #b ,illustration3 #c ,illustration3 #d #caf4--,,the ,,amazons4 ,it will be seen that only three queens have been removed from their positions on the edge of the board1 and that1 as a consequence1 eleven squares 7indicated by the black dots7 are left unattacked by any queen4 ,i will hazard the statement that eight queens cannot be placed on the chessboard so as to leave more than eleven squares unattacked4 ,it is true that we have no rigid proof of this yet1 but ,i have entirely convinced myself of the truth of the statement4 ,there are at least five different ways of arranging the queens so as to leave eleven squares unattacked4 ,illustration #cag4--;,a ,,puzzle ,,with ,,pawns4 ,sixteen pawns may be placed so that no three shall be in a straight line in any possible direction1 as in the diagram4 ,we regard1 as the conditions required1 the pawns as mere points on a plane4 ,illustration #cah4--,,lion-,,hunting4 ,there are #f1dhj ways of placing the man and the lion1 if there are no restrictions whatever except that they must be on different spots4 ,this is obvious1 because the man may be placed on any one of the #ha spots1 and in every case there are #hj spots remaining for the lion2 therefore #ha #hj #f1dhj4 ,now1 if we deduct the number of ways in which the lion and the man may be placed on the same path1 the result must be the number of ways in which they will not be on the same path4 ,the number of ways in which they may be in line is found without much difficulty to be #haf4 ,consequently1 #f1dhj - #haf #e1ffd1 the required answer4 ,the general solution is this3 #a_/cn7n - #a77cn - n #b74 ,this is1 of course1 equivalent to saying that if we call the number of squares on the side of a chessboard n1 then the formula shows the number of ways in which two bishops may be placed without attacking one another4 ,only in this case we must divide by two1 because the two bishops have no distinct individuality1 and cannot produce a different solution by mere exchange of places4 #cai4--,,the ,,knight-,,guards4 ,illustration3 ,,diagram #a4 ,illustration3 ,,diagram #b4 ,the smallest possible number of knights with which this puzzle can be solved is fourteen4 ,it has sometimes been assumed that there are a great many different solutions4 ,as a matter of fact1 there are only three arrangements--not counting mere reversals and reflections as different4 ,curiously enough1 nobody seems ever to have hit on the following simple proof1 or to have thought of dealing with the black and the white squares separately4 ,illustration3 ,,diagram #c4 ,illustration3 ,,diagram #d4 ,illustration3 ,,diagram #e4 ,seven knights can be placed on the board on white squares so as to attack every black square in two ways only4 ,these are shown in ,diagrams #a and #b4 ,note that three knights occupy the same position in both arrangements4 ,it is therefore clear that if we turn the board so that a black square shall be in the top left-hand corner instead of a white1 and place the knights in exactly the same positions1 we shall have two similar ways of attacking all the white squares4 ,i will assume the reader has made the two last described diagrams on transparent paper1 and marked them #a;a and #b;a4 ,now1 by placing the transparent ,diagram #a;a over #a you will be able to obtain the solution in ,diagram #c1 by placing #b;a over #b you will get ,diagram #d1 and by placing #b;a over #a you will get ,diagram #e4 ,you may now try all possible combinations of those two pairs of diagrams1 but you will only get the three arrangements ,i have given1 or their reversals and reflections4 ,therefore these three solutions are all that exist4 #cbj4--,,the ,,rook',s ,,tour4 ,illustration ,the only possible minimum solutions are shown in the two diagrams1 where it will be seen that only sixteen moves are required to perform the feat4 ,most people find it difficult to reduce the number of moves below seventeen994 ,illustration3 ,,the ,,rook',s ,,tour4 #cba4--,,the ,,rook',s ,,journey4 ,illustration ,i show the route in the diagram4 ,it will be seen that the tenth move lands us at the square marked #aj1 and that the last move1 the twenty-first1 brings us to a halt on square #ba4 #cbb4--,,the ,,languishing ,,maiden4 ,the dotted line shows the route in twenty-two straight paths by which the knight may rescue the maiden4 ,it is necessary1 after entering the first cell1 immediately to return before entering another4 ,otherwise a solution would not be possible4 7,see ,the ,grand ,tour1 p4 #bjj47 #cbc4--;,a ,,dungeon ,,puzzle4 ,if the prisoner takes the route shown in the diagram--where for clearness the doorways are omitted--he will succeed in visiting every cell once1 and only once1 in as many as fifty-seven straight lines4 ,no rook's path over the chessboard can exceed this number of moves4 ,illustration3 ,,the ,,languishing ,,maiden ,illustration3 ,a ,,dungeon ,,puzzle4 #cbd4--,,the ,,lion ,,and ,,the ,,man4 ,first of all1 the fewest possible straight lines in each case are twenty-two1 and in order that no cell may be visited twice it is absolutely necessary that each should pass into one cell and then immediately visit the one from which he started1 afterwards proceeding by way of the second available cell4 ,in the following diagram the man's route is indicated by the unbroken lines1 and the lion's by the dotted lines4 ,it will be found1 if the two routes are followed cell by cell with two pencil points1 that the lion and the man never meet4 ,but there was one little point that ought not to be overlooked--they occasionally got glimpses of one another4 ,now1 if we take one route for the man and merely reverse it for the lion1 we invariably find that1 going at the same speed1 they never get a glimpse of one another4 ,but in our diagram it will be found that the man and the lion are in the cells marked ,a at the same moment1 and may see one another through the open doorways2 while the same happens when they are in the two cells marked ,b1 the upper letters indicating the man and the lower the lion4 ,in the first case the lion goes straight for the man1 while the man appears to attempt to get in the rear of the lion2 in the second case it looks suspiciously like running away from one another6 ,illustration #cbe4--;,,an ,,episcopal ,,visitation4 ,illustration ,in the diagram ,i show how the bishop may be made to visit every one of his white parishes in seventeen moves4 ,it is obvious that we must start from one corner square and end at the one that is diagonally opposite to it4 ,the puzzle cannot be solved in fewer than seventeen moves4 #cbf4--;,a ,,new ,,counter ,,puzzle4 ,play as follows3 #b--c1 #i--d1 #aj--g1 #c--h1 #d--b1 #g--e1 #h--f1 #e--aj1 #f--i1 #b--e1 #a--f1 #f--d1 #e--c1 #aj--h1 #d--g1 #c--b1 #h--a1 #g--aj4 ,the white counters have now changed places with the red ones1 in eighteen moves1 without breaking the conditions4 #cbg4--;,a ,,new ,,bishop',s ,,puzzle4 ,illustration3 ,a ,illustration3 ,b ,play as follows1 using the notation indicated by the numbered squares in ,diagram ,a3-- ,white4 ,black4 ,white4 ,black4 #a4 #ah--ae #a4 #c--f #aj4 #bj--aj #aj4 #a--aa #b4 #ag--h #b4 #d--ac #aa4 #c--i #aa4 #ah--ab #c4 #ai--ad #c4 #b--g #ab4 #aj--ac #ab4 #aa--h #d4 #ae--e #d4 #f--af #ac4 #ai--af #ac4 #b--e #e4 #h--c #e4 #ac-ah #ad4 #af--a #ad4 #e--bj #f4 #ad--i #f4 #g--ab #ae4 #i--f #ae4 #ab--ae #g4 #e--aj #g4 #af-aa #af4 #ac-g #af4 #h--ad #h4 #i--ai #h4 #ab--b #ag4 #f--c #ag4 #ae-ah #i4 #aj--d #i4 #aa-ag #ah4 #g--b #ah4 #ad--ai ,diagram ,b shows the position after the ninth move4 ,bishops at #a and #bj have not yet moved1 but #b and #ai have sallied forth and returned4 ,in the end1 #a and #ai1 #b and #bj1 #c and #ag1 and #d and #ah will have exchanged places4 ,note the position after the thirteenth move4 #cbh4--,,the ,,queen',s ,,tour4 ,illustration ,the annexed diagram shows a second way of performing the ,queen's ,tour4 ,if you break the line at the point ,j and erase the shorter portion of that line1 you will have the required path solution for any ,j square4 ,if you break the line at ,i1 you will have a non-re-entrant solution starting from any ,i square4 ,and if you break the line at ,g1 you will have a solution for any ,g square4 ,the ,queen's ,tour previously given may be similarly broken at three different places1 but ,i seized the opportunity of exhibiting a second tour4 #cbi4--,,the ,,star ,,puzzle4 ,the illustration explains itself4 ,the stars are all struck out in fourteen straight strokes1 starting and ending at a white star4 ,illustration #ccj4--,,the ,,yacht ,,race4 ,the diagram explains itself4 ,the numbers will show the direction of the lines in their proper order1 and it will be seen that the seventh course ends at the flag-buoy1 as stipulated4 ,illustration #cca4--,,the ,,scientific ,,skater4 ,in this case we go beyond the boundary of the square4 ,apart from that1 the moves are all queen moves4 ,there are three or four ways in which it can be done4 ,here is one way of performing the feat3-- ,illustration ,it will be seen that the skater strikes out all the stars in one continuous journey of fourteen straight lines1 returning to the point from which he started4 ,to follow the skater's course in the diagram it is necessary always to go as far as we can in a straight line before turning4 #ccb4--,,the ,,forty-,,nine ,,stars4 ,the illustration shows how all the stars may be struck out in twelve straight strokes1 beginning and ending at a black star4 ,illustration #ccc4--,,the ,,queen',s ,,journey4 ,the correct solution to this puzzle is shown in the diagram by the dark line4 ,the five moves indicated will take the queen the greatest distance that it is possible for her to go in five moves1 within the conditions4 ,the dotted line shows the route that most people suggest1 but it is not quite so long as the other4 ,let us assume that the distance from the centre of any square to the centre of the next in the same horizontal or vertical line is #b inches1 and that the queen travels from the centre of her original square to the centre of the one at which she rests4 ,then the first route will be found to exceed #fg4i inches1 while the dotted route is less than #fg4h inches4 ,the difference is small1 but it is sufficient to settle the point as to the longer route4 ,all other routes are shorter still than these two4 ,illustration #ccd4--,,st4 ,,george ,,and ,,the ,,dragon4 ,we select for the solution of this puzzle one of the prettiest designs that can be formed by representing the moves of the knight by lines from square to square4 ,the chequering of the squares is omitted to give greater clearness4 ,st4 ,george thus slays the ,dragon in strict accordance with the conditions and in the elegant manner we should expect of him4 ,illustration3 ,st4 ,george and the ,dragon4 #cce4--;,,farmer ,,lawrence',s ,,cornfields4 ,there are numerous solutions to this little agricultural problem4 ,the version ,i give in the next column is rather curious on account of the long parallel straight lines formed by some of the moves4 ,illustration3 ,farmer ,lawrence's ,cornfields4 #ccf4--,,the ,,greyhound ,,puzzle4 ,there are several interesting points involved in this question4 ,in the first place1 if we had made no stipulation as to the positions of the two ends of the string1 it is quite impossible to form any such string unless we begin and end in the top and bottom row of kennels4 ,we may begin in the top row and end in the bottom 7or1 of course1 the reverse71 or we may begin in one of these rows and end in the same4 ,but we can never begin or end in one of the two central rows4 ,our places of starting and ending1 however1 were fixed for us4 ,yet the first half of our route must be confined entirely to those squares that are distinguished in the following diagram by circles1 and the second half will therefore be confined to the squares that are not circled4 ,the squares reserved for the two half-strings will be seen to be symmetrical and similar4 ,the next point is that the first half-string must end in one of the central rows1 and the second half-string must begin in one of these rows4 ,this is now obvious1 because they have to link together to form the complete string1 and every square on an outside row is connected by a knight's move with similar squares only--that is1 circled or non-circled as the case may be4 ,the half-strings can1 therefore1 only be linked in the two central rows4 ,illustration ,now1 there are just eight different first half-strings1 and consequently also eight second half-strings4 ,we shall see that these combine to form twelve complete strings1 which is the total number that exist and the correct solution of our puzzle4 ,i do not propose to give all the routes at length1 but ,i will so far indicate them that if the reader has dropped any he will be able to discover which they are and work them out for himself without any difficulty4 ,the following numbers apply to those in the above diagram4 ,the eight first half-strings are3 #a to #f 7#b routes72 #a to #h 7#a route72 #a to #aj 7#c routes72 #a to #ab 7#a route72 and #a to #ad 7#a route74 ,the eight second half-strings are3 #g to #bj 7#a route72 #i to #bj 7#a route72 #aa to #bj 7#c routes72 #ac to #bj 7#a route72 and #ae to #bj 7#b routes74 ,every different way in which you can link one half-string to another gives a different solution4 ,these linkings will be found to be as follows3 #f to #ac 7#b cases72 #aj to #ac 7#c cases72 #h to #aa 7#c cases72 #h to #ae 7#b cases72 #ab to #i 7#a case72 and #ad to #g 7#a case74 ,there are1 therefore1 twelve different linkings and twelve different answers to the puzzle4 ,the route given in the illustration with the greyhound will be found to consist of one of the three half-strings #a to #aj1 linked to the half-string #ac to #bj4 ,it should be noted that ten of the solutions are produced by five distinctive routes and their reversals--that is1 if you indicate these five routes by lines and then turn the diagrams upside down you will get the five other routes4 ,the remaining two solutions are symmetrical 7these are the cases where #ab to #i and #ad to #g are the links71 and consequently they do not produce new solutions by reversal4 #ccg4--,,the ,,four ,,kangaroos4 ,illustration ,a pretty symmetrical solution to this puzzle is shown in the diagram4 ,each of the four kangaroos makes his little excursion and returns to his corner1 without ever entering a square that has been visited by another kangaroo and without crossing the central line4 ,it will at once occur to the reader1 as a possible improvement of the puzzle1 to divide the board by a central vertical line and make the condition that this also shall not be crossed4 ,this would mean that each kangaroo had to confine himself to a square #d by #d1 but it would be quite impossible1 as ,i shall explain in the next two puzzles4 #cch4--,,the ,,board ,,in ,,compartments4 ,illustration ,in attempting to solve this problem it is first necessary to take the two distinctive compartments of twenty and twelve squares respectively and analyse them with a view to determining where the necessary points of entry and exit lie4 ,in the case of the larger compartment it will be found that to complete a tour of it we must begin and end on two of the outside squares on the long sides4 ,but though you may start at any one of these ten squares1 you are restricted as to those at which you can end1 or 7which is the same thing7 you may end at whichever of these you like1 provided you begin your tour at certain particular squares4 ,in the case of the smaller compartment you are compelled to begin and end at one of the six squares lying at the two narrow ends of the compartments1 but similar restrictions apply as in the other instance4 ,a very little thought will show that in the case of the two small compartments you must begin and finish at the ends that lie together1 and it then follows that the tours in the larger compartments must also start and end on the contiguous sides4 ,in the diagram given of one of the possible solutions it will be seen that there are eight places at which we may start this particular tour2 but there is only one route in each case1 because we must complete the compartment in which we find ourself before passing into another4 ,in any solution we shall find that the squares distinguished by stars must be entering or exit points1 but the law of reversals leaves us the option of making the other connections either at the diamonds or at the circles4 ,in the solution worked out the diamonds are used1 but other variations occur in which the circle squares are employed instead4 ,i think these remarks explain all the essential points in the puzzle1 which is distinctly instructive and interesting4 #cci4--,,the ,,four ,,knights' ,,tours4 ,illustration ,it will be seen in the illustration how a chessboard may be divided into four parts1 each of the same size and shape1 so that a complete re-entrant knight's tour may be made on each portion4 ,there is only one possible route for each knight and its reversal4 #cdj4--,,the ,,cubic ,,knight',s ,,tour4 ,illustration ,if the reader should cut out the above diagram1 fold it in the form of a cube1 and stick it together by the strips left for that purpose at the edges1 he would have an interesting little curiosity4 ,or he can make one on a larger scale for himself4 ,it will be found that if we imagine the cube to have a complete chessboard on each of its sides1 we may start with the knight on any one of the #chd squares1 and make a complete tour of the cube1 always returning to the starting-point4 ,the method of passing from one side of the cube to another is easily understood1 but1 of course1 the difficulty consisted in finding the proper points of entry and exit on each board1 the order in which the different boards should be taken1 and in getting arrangements that would comply with the required conditions4 #cda4--,,the ,,four ,,frogs4 ,the fewest possible moves1 counting every move separately1 are sixteen4 ,but the puzzle may be solved in seven plays1 as follows1 if any number of successive moves by one frog count as a single play4 ,all the moves contained within a bracket are a single play2 the numbers refer to the toadstools3 7#a--e71 7#c--g1 #g--a71 7#h--d1 #d--c1 #c--g71 7#f--b1 #b--h1 #h--d1 #d--c71 7#e--f1 #f--b1 #b--h71 7#a--e1 #e--f71 7#g--a74 ,this is the familiar old puzzle by ,guarini1 propounded in #aeab1 and ,i give it here in order to explain my buttons and string method of solving this class of moving-counter problem4 ,diagram ,a shows the old way of presenting ,guarini's puzzle1 the point being to make the white knights change places with the black ones4 ,in ,the ,four ,frogs presentation of the idea the possible directions of the moves are indicated by lines1 to obviate the necessity of the reader's understanding the nature of the knight's move in chess4 ,but it will at once be seen that the two problems are identical4 ,the central square can1 of course1 be ignored1 since no knight can ever enter it4 ,now1 regard the toadstools as buttons and the connecting lines as strings1 as in ,diagram ,b4 ,then by disentangling these strings we can clearly present the diagram in the form shown in ,diagram ,c1 where the relationship between the buttons is precisely the same as in ,b4 ,any solution on ,c will be applicable to ,b1 and to ,a4 ,place your white knights on #a and #c and your black knights on #f and #h in the ,c diagram1 and the simplicity of the solution will be very evident4 ,you have simply to move the knights round the circle in one direction or the other4 ,play over the moves given above1 and you will find that every little difficulty has disappeared4 ,illustrations3 ,a ,b ,c ,d ,e ,in ,diagram ,d ,i give another familiar puzzle that first appeared in a book published in ,brussels in #aghi1 ,les ,petites ,aventures de ,jerome ,sharp4 ,place seven counters on seven of the eight points in the following manner4 ,you must always touch a point that is vacant with a counter1 and then move it along a straight line leading from that point to the next vacant point 7in either direction71 where you deposit the counter4 ,you proceed in the same way until all the counters are placed4 ,remember you always touch a vacant place and slide the counter from it to the next place1 which must be also vacant4 ,now1 by the buttons and string method of simplification we can transform the diagram into ,e4 ,then the solution becomes obvious4 ,always move to the point that you last moved from4 ,this is not1 of course1 the only way of placing the counters1 but it is the simplest solution to carry in the mind4 ,there are several puzzles in this book that the reader will find lend themselves readily to this method4 #cdb4--,,the ,,mandarin',s ,,puzzle4 ,the rather perplexing point that the solver has to decide for himself in attacking this puzzle is whether the shaded numbers 7those that are shown in their right places7 are mere dummies or not4 ,ninety-nine persons out of a hundred might form the opinion that there can be no advantage in moving any of them1 but if so they would be wrong4 ,the shortest solution without moving any shaded number is in thirty-two moves4 ,but the puzzle can be solved in thirty moves4 ,the trick lies in moving the #f1 or the #ae1 on the second move and replacing it on the nineteenth move4 ,here is the solution3 #b1 #f1 #ac1 #d1 #a1 #ba1 #d1 #a1 #aj1 #b1 #ba1 #aj1 #b1 #e1 #bb1 #af1 #a1 #ac1 #f1 #ai1 #aa1 #b1 #e1 #bb1 #af1 #e1 #ac1 #d1 #aj1 #ba4 ,thirty moves4 #cdc4--;,,exercise ,,for ,,prisoners4 ,there are eighty different arrangements of the numbers in the form of a perfect knight's path1 but only forty of these can be reached without two men ever being in a cell at the same time4 ,two is the greatest number of men that can be given a complete rest1 and though the knight's path can be arranged so as to leave either #g and #ac1 #h and #ac1 #e and #g1 or #e and #ac in their original positions1 the following four arrangements1 in which #g and #ac are unmoved1 are the only ones that can be reached under the moving conditions4 ,it therefore resolves itself into finding the fewest possible moves that will lead up to one of these positions4 ,this is certainly no easy matter1 and no rigid rules can be laid down for arriving at the correct answer4 ,it is largely a matter for individual judgment1 patient experiment1 and a sharp eye for revolutions and position4 ,a -------- #f #aajae -------- #iab #g #d -------- #b #eadaa -------- #ac #h #c9999 -------- ,b -------- #f #aajae -------- #aaad #g #d -------- #b #eab #i -------- #ac #h #c9999 -------- ,c -------- #f #i #dae -------- #aab #gaj -------- #h #ead #c -------- #ac #baa9999 -------- ,d -------- #faa #dae -------- #aad #gaj -------- #h #eab #c -------- #ac #b #i9999 -------- ,illustration3 ,a1 ,b1 ,c1 ,d ,as a matter of fact1 the position ,c can be reached in as few as sixty-six moves in the following manner3 #ab1 #aa1 #ae1 #ab1 #aa1 #h1 #d1 #c1 #b1 #f1 #e1 #a1 #f1 #e1 #aj1 #ae1 #h1 #d1 #c1 #b1 #e1 #aj1 #ae1 #h1 #d1 #c1 #b1 #e1 #aj1 #ae1 #h1 #d1 #ab1 #aa1 #c1 #b1 #e1 #aj1 #ae1 #f1 #a1 #h1 #d1 #i1 #h1 #a1 #f1 #d1 #i1 #ab1 #b1 #e1 #aj1 #ae1 #d1 #i1 #ab1 #b1 #e1 #c1 #aa1 #ad1 #b1 #e1 #ad1 #aa #ff moves4 ,though this is the shortest that ,i know of1 and ,i do not think it can be beaten1 ,i cannot state positively that there is not a shorter way yet to be discovered4 ,the most tempting arrangement is certainly ,a2 but things are not what they seem1 and ,c is really the easiest to reach4 ,if the bottom left-hand corner cell might be left vacant1 the following is a solution in forty-five moves by ,mr4 ,r4 ,elrick3 #ae1 #aa1 #aj1 #i1 #ac1 #ad1 #aa1 #aj1 #g1 #h1 #d1 #c1 #h1 #f1 #i1 #g1 #ab1 #d1 #f1 #i1 #e1 #ac1 #g1 #e1 #ac1 #a1 #b1 #ac1 #e1 #g1 #a1 #b1 #ac1 #h1 #c1 #f1 #i1 #ab1 #g1 #aa1 #ad1 #a1 #aa1 #ad1 #a4 ,but every man has moved4 #cdd4--,,the ,,kennel ,,puzzle4 ,the first point is to make a choice of the most promising knight's string and then consider the question of reaching the arrangement in the fewest moves4 ,i am strongly of opinion that the best string is the one represented in the following diagram1 in which it will be seen that each successive number is a knight's move from the preceding one1 and that five of the dogs 7#a1 #e1 #aj1 #ae1 and #bj7 never leave their original kennels4 ----------------------------- #a #b #c #d #e #a #ah #i #ad #e ----------------------------- #f #g #h #i #aj #h #ac #d #ai #aj ----------------------------- #aa #ab #ac #ad #ae #ag #b #aa #f #ae ----------------------------- #af #ag #ah #ai #bj #ab #g #af #c #bj ----------------------------- #ba #bb #bc #bd #be ----------------------------- ,illustration ,this position may be arrived at in as few as forty-six moves1 as follows3 #af--ba1 #af--bb1 #af--bc1 #ag--af1 #ab--ag1 #ab--bb1 #ab--ba1g--ab1 #g--ag1 #g--bb1 #aa--ab1 #aa--ag1 #b--g1 #b--ab1 #f--aa1 #h--g1 #h--f1 #ac--h1 #ah--ac1 #aa--ah1 #b--ag1 #ah--ab1 #ah--g1 #ah--b1 #ac--g1 #c--h1 #c--ac1 #d--c1 #d--h1 #i--d1 #i--c1 #ad--i1 #ad--d1 #ai--ad1 #ai--i1 #c--ad1 #c--ai1 #f--ab1 #f--ac1 #f--ad1 #ag--aa1 #ab--af1 #b--ab1 #g--ag1 #aa--ac1 #af--ah #df moves4 ,i am1 of course1 not able to say positively that a solution cannot be discovered in fewer moves1 but ,i believe it will be found a very hard task to reduce the number4 #cde4--,,the ,,two ,,pawns4 ,call one pawn ,a and the other ,b4 ,now1 owing to that optional first move1 either pawn may make either #e or #f moves in reaching the eighth square4 ,there are1 therefore1 four cases to be considered3 7#a7 ,a #f moves and ,b #f moves2 7#b7 ,a #f moves and ,b #e moves2 7#c7 ,a #e moves and ,b #f moves2 7#d7 ,a #e moves and ,b #e moves4 ,in case 7#a7 there are #ab moves1 and we may select any #f of these for ,a4 ,therefore #ghiajaaab divided by #abcdef gives us the number of variations for this case--that is1 #ibd4 ,similarly for case 7#b71 #f selections out of #aa will be #dfb2 in case 7#c71 #e selections out of #aa will also be #dfb2 and in case 7#d71 #e selections out of #aj will be #beb4 ,add these four numbers together and we get #b1ajj1 which is the correct number of different ways in which the pawns may advance under the conditions4 7,see ,no4 #bgj1 on p4 #bjd47 #cdf4--,,setting ,,the ,,board4 ,the ,white pawns may be arranged in #dj1cbj ways1 the ,white rooks in #b ways1 the bishops in #b ways1 and the knights in #b ways4 ,multiply these numbers together1 and we find that the ,white pieces may be placed in #cbb1efj different ways4 ,the ,black pieces may1 of course1 be placed in the same number of ways4 ,therefore the men may be set up in #cbb1efj #cbb1efj #ajd1jdd1iec1fjj ways4 ,but the point that nearly everybody overlooks is that the board may be placed in two different ways for every arrangement4 ,therefore the answer is doubled1 and is #bjh1jhi1ijg1bjj different ways4 #cdg4--;,,counting ,,the ,,rectangles4 ,there are #a1bif different rectangles in all1 #bjd of which are squares1 counting the square board itself as one1 and #a1jib rectangles that are not squares4 ,the general formula is that a board of n squares contains 77n n77_/#d rectangles1 of which 7#bn #cn n7_/#f are squares and 7#cn#d #bn - #cn - #bn7_/#ab are rectangles that are not squares4 ,it is curious and interesting that the total number of rectangles is always the square of the triangular number whose side is n4 #cdh4--,,the ,,rookery4 ,the answer involves the little point that in the final position the numbered rooks must be in numerical order in the direction contrary to that in which they appear in the original diagram1 otherwise it cannot be solved4 ,play the rooks in the following order of their numbers4 ,as there is never more than one square to which a rook can move 7except on the final move71 the notation is obvious--#e1 #f1 #g1 #e1 #f1 #d1 #c1 #f1 #d1 #g1 #e1 #d1 #g1 #c1 #f1 #g1 #c1 #e1 #d1 #c1 #a1 #h1 #c1 #d1 #e1 #f1 #g1 #a1 #h1 #b1 #a1 and rook takes bishop1 checkmate4 ,these are the fewest possible moves--thirty-two4 ,the ,black king's moves are all forced1 and need not be given4 #cdi4--,,stalemate4 ,working independently1 the same position was arrived at by ,messrs4 ,s4 ,loyd1 ,e4,n4 ,frankenstein1 ,w4,h4 ,thompson1 and myself4 ,so the following may be accepted as the best solution possible to this curious problem 3-- ,white4 ,black4 #a4 ,p--,q#d #a4 ,p--,k#d #b4 ,q--,q#c #b4 ,q--,r#e #c4 ,q--,k,kt#c #c4 ,b--,kt#e ch #d4 ,kt--,q#b #d4 ,p--,,qr#d #e4 ,p--,r#d #e4 ,p--,q#c #f4 ,p--,r#c #f4 ,b--,k#c #g4 ,r--,r#c #g4 ,p--,,kb#d #h4 ,q--,r#b #h4 ,p--,b#d #i4 ,r--,k,kt#c #i4 ,b--,kt#f #aj4 ,p--,,qb#d #aj4 ,p--,b#e #aa4 ,p--,b#c #aa4 ,p--,k#e #ab4 ,p--,q#e #ab4 ,p--,k#f ,and ,white is stalemated4 ,we give a diagram of the curious position arrived at4 ,it will be seen that not one of ,white's pieces may be moved4 ,illustration -------- rn k nr -------- p pp -------- p -------- p p,p -------- ,pb,p p q -------- b p,p,r,p -------- ,p ,n,p ,p,q -------- ,b ,k,b,n,r -------- #cej4--,,the ,,forsaken ,,king4 ,play as follows3-- ,white4 ,black4 #a4 ,p to ,k #dth #a4 ,any move #b4 ,q to ,kt #dth #b4 ,any move except on ,,kb file 7a7 #c4 ,q to ,kt #gth #c4 ,k moves to royal row #d4 ,b to ,kt #eth #d4 ,any move #e4 ,mate in two moves ,if #c4 ,k other than to royal row #d4 ,p to ,q #dth #d4 ,any move #e4 ,mate in two moves 7a7 ,if #b4 ,any move on ,,kb file #c4 ,q to ,q #gth #c4 ,k moves to royal row #d4 ,p to ,q ,kt #crd #d4 ,any move #e4 ,mate in two moves ,if #c4 ,k other than to royal row #d4 ,p to ,q #dth #d4 ,any move #e4 ,mate in two moves ,of course1 by royal row is meant the row on which the king originally stands at the beginning of a game4 ,though1 if ,black plays badly1 he may1 in certain positions1 be mated in fewer moves1 the above provides for every variation he can possibly bring about4 #cea4--,,the ,,crusader4 ,white4 ,black4 #a4 ,kt to ,,qb #crd #a4 ,p to ,q #dth #b4 ,kt takes ,,qp #b4 ,kt to ,,qb #crd #c4 ,kt takes ,,kp #c4 ,p to ,k,kt #dth #d4 ,kt takes ,b #d4 ,kt to ,,kb #crd #e4 ,kt takes ,p #e4 ,kt to ,k #eth #f4 ,kt takes ,kt #f4 ,kt to ,b #fth #g4 ,kt takes ,q #g4 ,r to ,k,kt sq #h4 ,kt takes ,,bp #h4 ,r to ,k,kt #crd #i4 ,kt takes ,p #i4 ,r to ,k #crd #aj4 ,kt takes ,p #aj4 ,kt to ,kt #hth #aa4 ,kt takes ,b #aa4 ,r to ,r #fth #ab4 ,kt takes ,r #ab4 ,p to ,kt #dth #ac4 ,kt takes ,p 7ch7 #ac4 ,k to ,b #bnd #ad4 ,kt takes ,p #ad4 ,k to ,kt #crd #ae4 ,kt takes ,r #ae4 ,k to ,r #dth #af4 ,kt takes ,kt #af4 ,k to ,r #eth ,white now mates in three moves4 #ag4 ,p to ,q #dth #ag4 ,k to ,r #dth #ah4 ,q to ,q #crd #ah4 ,k moves #ai4 ,q to ,,kr #crd 7mate7 ,if #ag4 ,k to ,kt #eth #ah4 ,p to ,k #dth 7dis4 ch7 #ah4 ,k moves #ai4 ,p to ,k,kt #crd 7mate7 ,the position after the sixteenth move1 with the mate in three moves1 was first given by ,s4 ,loyd in ,chess ,nuts4 #ceb4--;,,immovable ,,pawns4 #a4 ,kt to ,,kb #c #b4 ,kt to ,,kr #d #c4 ,kt to ,kt #f #d4 ,kt takes ,r #e4 ,kt to ,kt #f #f4 ,kt takes ,b #g4 ,k takes ,kt #h4 ,kt to ,,qb #c #i4 ,kt to ,r #d #aj4 ,kt to ,kt #f #aa4 ,kt takes ,r #ab4 ,kt to ,kt #f #ac4 ,kt takes ,b #ad4 ,kt to ,q #f #ae4 ,q to ,k sq #af4 ,kt takes ,q #ag4 ,k takes ,kt1 and the position is reached4 ,black plays precisely the same moves as ,white1 and therefore we give one set of moves only4 ,the above seventeen moves are the fewest possible4 #cec4--,,thirty-,,six ,,mates4 ,place the remaining eight ,white pieces thus3 ,k at ,,kb #dth1 ,q at ,q,kt #fth1 ,r at ,q #fth1 ,r at ,k,kt #gth1 ,b at ,q #eth1 ,b at ,,kr #hth1 ,kt at ,,qr #eth1 and ,kt at ,,qb #eth4 ,the following mates can then be given3-- ,by discovery from ,q #h ,by discovery from ,r at ,q #fth #ac ,by discovery from ,b at ,r #hth #aa ,given by ,kt at ,r #eth #b ,given by pawns #b -- ,total #cf ,is it possible to construct a position in which more than thirty-six different mates on the move can be given8 ,so far as ,i know1 nobody has yet beaten my arrangement4 #ced4--;,,an ,,amazing ,,dilemma4 ,mr ,black left his king on his queen's knight's #gth1 and no matter what piece ,white chooses for his pawn1 ,black cannot be checkmated4 ,as we said1 the ,black king takes no notice of checks and never moves4 ,white may queen his pawn1 capture the ,black rook1 and bring his three pieces up to the attack1 but mate is quite impossible4 ,the ,black king cannot be left on any other square without a checkmate being possible4 ,the late ,sam ,loyd first pointed out the peculiarity on which this puzzle is based4 #cee4--;,,checkmate6 ,remove the ,white pawn from ,b #fth to ,k #dth and place a ,black pawn on ,black's ,,kb #bnd4 ,now1 ,white plays ,p to ,k #eth1 check1 and ,black must play ,p to ,b #dth4 ,then ,white plays ,p takes ,p en passant1 checkmate4 ,this was therefore ,white's last move1 and leaves the position given4 ,it is the only possible solution4 #cef4--,,queer ,,chess4 -------- -------- ,rk,r,n -------- -------- ,if you place the pieces as follows 7where only a portion of the board is given1 to save space71 the ,black king is in check1 with no possible move open to him4 ,the reader will now see why ,i avoided the term checkmate1 apart from the fact that there is no ,white king4 ,the position is impossible in the game of chess1 because ,black could not be given check by both rooks at the same time1 nor could he have moved into check on his last move4 ,i believe the position was first published by the late ,s4 ,loyd4 #ceg4--;,,ancient ,,chinese ,,puzzle4 ,play as follows3-- #a4 ,r--,q #f #b4 ,k--,r #g #c4 ,r 7,r #f7--;,b #f 7mate74 ,black's moves are forced1 so need not be given4 #ceh4--,,the ,,six ,,pawns4 ,the general formula for six pawns on all squares greater than #b is this3 ,six times the square of the number of combinations of n things taken three at a time1 where n represents the number of squares on the side of the board4 ,of course1 where n is even the unoccupied squares in the rows and columns will be even1 and where n is odd the number of squares will be odd4 ,here n is #h1 so the answer is #ah1haf different ways4 ,this is ,the ,dyer's ,puzzle 7,canterbury ,puzzles1 ,no4 #bg7 in another form4 ,i repeat it here in order to explain a method of solving that will be readily grasped by the novice4 ,first of all1 it is evident that if we put a pawn on any line1 we must put a second one in that line in order that the remainder may be even in number4 ,we cannot put four or six in any row without making it impossible to get an even number in all the columns interfered with4 ,we have1 therefore1 to put two pawns in each of three rows and in each of three columns4 ,now1 there are just six schemes or arrangements that fulfil these conditions1 and these are shown in ,diagrams ,a to ,f1 inclusive1 on next page4 ,illustration ,i will just remark in passing that ,a and ,b are the only distinctive arrangements1 because1 if you give ,a a quarter-turn1 you get ,f2 and if you give ,b three quarter-turns in the direction that a clock hand moves1 you will get successively ,c1 ,d1 and ,e4 ,no matter how you may place your six pawns1 if you have complied with the conditions of the puzzle they will fall under one of these arrangements4 ,of course it will be understood that mere expansions do not destroy the essential character of the arrangements4 ,thus ,g is only an expansion of form ,a4 ,the solution therefore consists in finding the number of these expansions4 ,supposing we confine our operations to the first three rows1 as in ,g1 then with the pairs a and b placed in the first and second columns the pair c may be disposed in any one of the remaining six columns1 and so give six solutions4 ,now slide pair b into the third column1 and there are five possible positions for c4 ,slide b into the fourth column1 and c may produce four new solutions4 ,and so on1 until 7still leaving a in the first column7 you have b in the seventh column1 and there is only one place for c--in the eighth column4 ,then you may put a in the second column1 b in the third1 and c in the fourth1 and start sliding c and b as before for another series of solutions4 ,we find thus that1 by using form ,a alone and confining our operations to the three top rows1 we get as many answers as there are combinations of #h things taken #c at a time4 ,this is 7#h #g #f7_/7a #b #c7 #ef4 ,and it will at once strike the reader that if there are #ef different ways of electing the columns1 there must be for each of these ways just #ef ways of selecting the rows1 for we may simultaneously work that sliding process downwards to the very bottom in exactly the same way as we have worked from left to right4 ,therefore the total number of ways in which form ,a may be applied is #ef #f #c1acf4 ,but there are1 as we have seen1 six arrangements1 and we have only dealt with one of these1 ,a4 ,we must1 therefore1 multiply this result by #f1 which gives us #c1acf #f #ah1haf1 which is the total number of ways1 as we have already stated4 #cei4--;,,counter ,,solitaire4 ,play as follows3 #c--aa1 #i--aj1 #a--b1 #g--ae1 #h--af1 #h--g1 #e--ac1 #a--d1 #h--e1 #f--ad1 #c--h1 #f--c1 #f--ab1 #a--f1 #a--i1 and all the counters will have been removed1 with the exception of ,no4 #a1 as required by the conditions4 #cfj4--;,,chessboard ,,solitaire4 ,play as follows3 #g--ae1 #h--af1 #h--g1 #b--aj1 #a--i1 #a--b1 #e--ac1 #c--d1 #f--c1 #aa--a1 #ad--h1 #f--ab1 #e--f1 #e--aa1 #ca--bc1 #cb--bd1 #cb--ca1 #bf--ah1 #be--ag1 #be--bf1 #bb--cb1 #ad--bb1 #bi--ba1 #ad--bi1 #bg--bh1 #cj--bg1 #be--ad1 #cj--bj1 #be--cj1 #be--e4 ,the two counters left on the board are #be and #ai--;both belonging to the same group1 as stipulated--and #ai has never been moved from its original place4 ,i do not think any solution is possible in which only one counter is left on the board4 #cfa4--,,the ,,monstrosity4 ,white ,black1 #a4 ,p to ,,kb #d ,p to ,,qb #c #b4 ,k to ,b #b ,q to ,r #d #c4 ,k to ,k #c ,k to ,q sq #d4 ,p to ,b #e ,k to ,b #b #e4 ,q to ,k sq ,k to ,kt #c #f4 ,q to ,kt #c ,kt to ,,qr #c #g4 ,q to ,kt #h ,p to ,,kr #d #h4 ,kt to ,,kb #c ,r to ,r #c #i4 ,kt to ,k #e ,r to ,kt #c #aj4 ,q takes ,b ,r to ,kt #f1 ch #aa4 ,p takes ,r ,k to ,kt #d #ab4 ,r to ,r #d ,p to ,b #c #ac4 ,r to ,q #d ,p takes ,kt #ad4 ,p to ,q,kt #d ,p takes ,r1 ch #ae4 ,k to ,b #d ,p to ,r #e #af4 ,q to ,k #h ,p to ,r #f #ag4 ,kt to ,b #c1 ch ,p takes ,kt #ah4 ,b to ,r #c ,p to ,r #g #ai4 ,r to ,kt sq ,p to ,r #h 7,q7 #bj4 ,r to ,kt #b ,p takes ,r #ba4 ,k to ,kt #e ,q to ,k,kt #h #bb4 ,q to ,r #e ,k to ,r #e #bc4 ,p to ,kt #e ,r to ,b sq #bd4 ,p to ,kt #f ,r to ,b #b #be4 ,p takes ,r ,p to ,kt #h 7,b7 #bf4 ,p to ,b #h 7,r7 ,q to ,b #b #bg4 ,b to ,q #f ,kt to ,kt #e #bh4 ,k to ,kt #f ,k to ,r #f #bi4 ,r to ,r #h ,k to ,kt #g #cj4 ,p to ,r #d ,q 7,kt #h7 to ,kt #c #ca4 ,p to ,r #e ,k to ,b #h #cb4 ,p takes ,q ,k to ,q #h #cc4 ,p takes ,q ,k to ,k #h #cd4 ,k to ,b #g ,kt to ,,kr #c1 ch #ce4 ,k to ,k #h ,b to ,r #g #cf4 ,p to ,b #f ,b to ,kt sq #cg4 ,p to ,b #g ,k takes ,b #ch4 ,p to ,b #h 7,b7 ,kt to ,q #d #ci4 ,b to ,kt #h ,kt to ,b #c1 ch #dj4 ,k to ,q #h ,kt to ,k sq #da4 ,p takes ,kt 7,r7 ,kt to ,b #b1 ch #db4 ,k to ,b #g ,kt to ,q sq #dc4 ,q to ,b #g1 ch ,k to ,kt #h ,and the position is reached4 ,the order of the moves is immaterial1 and this order may be greatly varied4 ,but1 although many attempts have been made1 nobody has succeeded in reducing the number of my moves4 #cfb4--,,the ,,wassail ,,bowl4 ,the division of the twelve pints of ale can be made in eleven manipulations1 as below4 ,the six columns show at a glance the quantity of ale in the barrel1 the five-pint jug1 the three-pint jug1 and the tramps ,x1 ,y1 and ,z respectively after each manipulation4 ,barrel4 #e-pint4 #c-pint4 ,x4 ,y4 ,z4 #g 44 #e 44 #j 44 #j 44 #j 44 #j #g 44 #b 44 #c 44 #j 44 #j 44 #j #g 44 #j 44 #c 44 #b 44 #j 44 #j #g 44 #c 44 #j 44 #b 44 #j 44 #j #d 44 #c 44 #c 44 #b 44 #j 44 #j #j 44 #c 44 #c 44 #b 44 #d 44 #j #j 44 #e 44 #a 44 #b 44 #d 44 #j #j 44 #e 44 #j 44 #b 44 #d 44 #a #j 44 #b 44 #c 44 #b 44 #d 44 #a #j 44 #j 44 #c 44 #d 44 #d 44 #a #j 44 #j 44 #j 44 #d 44 #d 44 #d ,and each man has received his four pints of ale4 #cfc4--,,the ,,doctor',s ,,query4 ,the mixture of spirits of wine and water is in the proportion of #dj to #a1 just as in the other bottle it was in the proportion of #a to #dj4 #cfd4--,,the ,,barrel ,,puzzle4 ,illustration3 ,figs4 #a1 #b1 and #c ,all that is necessary is to tilt the barrel as in ,fig4 #a1 and if the edge of the surface of the water exactly touches the lip a at the same time that it touches the edge of the bottom b1 it will be just half full4 ,to be more exact1 if the bottom is an inch or so from the ground1 then we can allow for that1 and the thickness of the bottom1 at the top4 ,if when the surface of the water reached the lip a it had risen to the point c in ,fig4 #b1 then it would be more than half full4 ,if1 as in ,fig4 #c1 some portion of the bottom were visible and the level of the water fell to the point d1 then it would be less than half full4 ,this method applies to all symmetrically constructed vessels4 #cfe4--,,new ,,measuring ,,puzzle4 ,the following solution in eleven manipulations shows the contents of every vessel at the start and after every manipulation3-- #aj-quart4 #aj-quart4 #e-quart4 #d-quart4 #aj 44 #aj 44 #j 44 #j #e 44 #aj 44 #e 44 #j #e 44 #aj 44 #a 44 #d #i 44 #aj 44 #a 44 #j #i 44 #f 44 #a 44 #d #i 44 #g 44 #j 44 #d #i 44 #g 44 #d 44 #j #i 44 #c 44 #d 44 #d #i 44 #c 44 #e 44 #c #i 44 #h 44 #j 44 #c #d 44 #h 44 #e 44 #c #d 44 #aj 44 #c 44 #c #cff4--,,the ,,honest ,,dairyman4 ,whatever the respective quantities of milk and water1 the relative proportion sent to ,london would always be three parts of water to one of milk4 ,but there are one or two points to be observed4 ,there must originally be more water than milk1 or there will be no water in ,a to double in the second transaction4 ,and the water must not be more than three times the quantity of milk1 or there will not be enough liquid in ,b to effect the second transaction4 ,the third transaction has no effect on ,a1 as the relative proportions in it must be the same as after the second transaction4 ,it was introduced to prevent a quibble if the quantity of milk and water were originally the same2 for though double nothing would be nothing1 yet the third transaction in such a case could not take place4 #cfg4--,,wine ,,and ,,water4 ,the wine in small glass was one-sixth of the total liquid1 and the wine in large glass two-ninths of total4 ,add these together1 and we find that the wine was seven-eighteenths of total fluid1 and therefore the water eleven-eighteenths4 #cfh4--,,the ,,keg ,,of ,,wine4 ,the capacity of the jug must have been a little less than three gallons4 ,to be more exact1 it was #b4ic gallons4 #cfi4--,,mixing ,,the ,,tea4 ,there are three ways of mixing the teas4 ,taking them in the order of quality1 #bs4 #f;d41 #bs4 #c;d41 #as4 #ip41 mix #af lbs41 #a lb41 #c lbs42 or #ad lbs41 #d lbs41 #b lbs42 or #ab lbs41 #g lbs41 #a lb4 ,in every case the twenty pounds mixture should be worth #bs4 #d;d4 per pound2 but the last case requires the smallest quantity of the best tea1 therefore it is the correct answer4 #cgj4--;,a ,,packing ,,puzzle4 ,on the side of the box1 #ad by #bbd_/e1 we can arrange #ac rows containing alternately #g and #f balls1 or #he in all4 ,above this we can place another layer consisting of #ab rows of #g and #f alternately1 or a total of #gh4 ,in the length of #bdi_/aj inches #ae such layers may be packed1 the alternate layers containing #he and #gh balls4 ,thus #h times #he added to #g times #gh gives us #a1bbf for the full contents of the box4 #cga4--;,,gold ,,packing ,,in ,,russia4 ,the box should be #ajj inches by #ajj inches by #aa inches deep1 internal dimensions4 ,we can lay flat at the bottom a row of eight slabs1 lengthways1 end to end1 which will just fill one side1 and nine of these rows will dispose of seventy-two slabs 7all on the bottom71 with a space left over on the bottom measuring #ajj inches by #a inch by #a inch4 ,now make eleven depths of such seventy-two slabs1 and we have packed #gib1 and have a space #ajj inches by #a inch by #aa inches deep4 ,in this we may exactly pack the remaining eight slabs on edge1 end to end4 #cgb4--,,the ,,barrels ,,of ,,honey4 ,the only way in which the barrels could be equally divided among the three brothers1 so that each should receive his #c barrels of honey and his #g barrels1 is as follows3-- ,full4 ,half-full4 ,empty4 ,a #c #a #c ,b #b #c #b ,c #b #c #b ,there is one other way in which the division could be made1 were it not for the objection that all the brothers made to taking more than four barrels of the same description4 ,except for this difficulty1 they might have given ,b his quantity in exactly the same way as ,a above1 and then have left ,c one full barrel1 five half-full barrels1 and one empty barrel4 ,it will thus be seen that in any case two brothers would have to receive their allowance in the same way4 #cgc4--;,,crossing ,,the ,,stream4 ,first1 the two sons cross1 and one returns ,then the man crosses and the other son returns4 ,then both sons cross and one returns4 ,then the lady crosses and the other son returns ,then the two sons cross and one of them returns for the dog4 ,eleven crossings in all4 ,it would appear that no general rule can be given for solving these river-crossing puzzles4 ,a formula can be found for a particular case 7say on ,no4 #cge or #cgf7 that would apply to any number of individuals under the restricted conditions2 but it is not of much use1 for some little added stipulation will entirely upset it4 ,as in the case of the measuring puzzles1 we generally have to rely on individual ingenuity4 #cgd4--;,,crossing ,,the ,,river ,,axe4 ,here is the solution3-- ,j #e7 ,g ,t#h #c #e 7 ,j ,g ,t#h #c #e ,g #c7 ,,jt#h #ec 7 ,g ,,jt#h #ec ,j ,t7 ,g #h ,j #e 7,t #c ,g #h ,j #e ,g #h7 ,t #c ,g #h 7,j #e ,t ,g #h ,j ,t7 #ec ,,jt#h 7 ,g #ec ,,jt#h ,g #c7 #e ,g ,t#h #c 7 ,j #e ,g ,t#h #c ,j #e7 ,g1 ,j1 and ,t stand for ,giles1 ,jasper1 and ,timothy2 and #h1 #e1 #c1 for #hjj1 #ejj1 and #cjj respectively4 ,the two side columns represent the left bank and the right bank1 and the middle column the river4 ,thirteen crossings are necessary1 and each line shows the position when the boat is in mid-stream during a crossing1 the point of the bracket indicating the direction4 ,it will be found that not only is no person left alone on the land or in the boat with more than his share of the spoil1 but that also no two persons are left with more than their joint shares1 though this last point was not insisted upon in the conditions4 #cge4--;,,five ,,jealous ,,husbands4 ,it is obvious that there must be an odd number of crossings1 and that if the five husbands had not been jealous of one another the party might have all got over in nine crossings4 ,but no wife was to be in the company of a man or men unless her husband was present4 ,this entails two more crossings1 eleven in all4 ,the following shows how it might have been done4 ,the capital letters stand for the husbands1 and the small letters for their respective wives4 ,the position of affairs is shown at the start1 and after each crossing between the left bank and the right1 and the boat is represented by the asterisk4 ,so you can see at a glance that a1 b1 and c went over at the first crossing1 that b and c returned at the second crossing1 and so on4 ,,abcde abcde 9944 #a4 ,,abcde de 4499 abc #b4 ,,abcde bcde 9944 a #c4 ,,abcde e 4499 abcd #d4 ,,abcde de 9944 abc #e4 ,,de de 1199 ,,abc abc #f4 ,,cde cde 9944 ,,ab ab #g4 cde 4499 ,,abcde ab #h4 bcde 9944 ,,abcde a #i4 e 4499 ,,abcde abcd #aj4 bc e 9944 ,,abcde a d #aa4 4499 ,,abcde abcde ,there is a little subtlety concealed in the words show the quickest way4 ,everybody correctly assumes that1 as we are told nothing of the rowing capabilities of the party1 we must take it that they all row equally well4 ,but it is obvious that two such persons should row more quickly than one4 ,therefore in the second and third crossings two of the ladies should take back the boat to fetch d1 not one of them only4 ,this does not affect the number of landings1 so no time is lost on that account4 ,a similar opportunity occurs in crossings #aj and #aa1 where the party again had the option of sending over two ladies or one only4 ,to those who think they have solved the puzzle in nine crossings ,i would say that in every case they will find that they are wrong4 ,no such jealous husband would1 in the circumstances1 send his wife over to the other bank to a man or men1 even if she assured him that she was coming back next time in the boat4 ,if readers will have this fact in mind1 they will at once discover their errors4 #cgf4--,,the ,,four ,,elopements4 ,if there had been only three couples1 the island might have been dispensed with1 but with four or more couples it is absolutely necessary in order to cross under the conditions laid down4 ,it can be done in seventeen passages from land to land 7though ,french mathematicians have declared in their books that in such circumstances twenty-four are needed71 and it cannot be done in fewer4 ,i will give one way4 ,a1 ,b1 ,c1 and ,d are the young men1 and a1 b1 c1 and d are the girls to whom they are respectively engaged4 ,the three columns show the positions of the different individuals on the lawn1 the island1 and the opposite shore before starting and after each passage1 while the asterisk indicates the position of the boat on every occasion4 ,lawn4 ,island4 ,shore4 ,a,b,c,dabcd 99 ,,abcd cd ab 99 ,,abcd bcd 99 a ,,abcd d bc 99 a ,,abcd cd 99 b a ,,cd cd b ,,ab a 99 ,,bcd cd 99 b ,a a ,,bcd bcd 99 ,a a ,,bcd d 99 bc ,a a ,d d bc ,,abc a 99 ,d d abc 99 ,,abc ,d d b ,,abc a c 99 ,b ,d d 99 b ,a ,c a c d b ,,abcd a c 99 d bc 99 ,,abcd a d ,,abcd abc 99 cd 99 ,,abcd ab ,,abcd abcd 99 ,having found the fewest possible passages1 we should consider two other points in deciding on the quickest method3 ,which persons were the most expert in handling the oars1 and which method entails the fewest possible delays in getting in and out of the boat8 ,we have no data upon which to decide the first point1 though it is probable that1 as the boat belonged to the girls' household1 they would be capable oarswomen4 ,the other point1 however1 is important1 and in the solution ,i have given 7where the girls do #h-acths of the rowing and ,a and ,d need not row at all7 there are only sixteen gettings-in and sixteen gettings-out4 ,a man and a girl are never in the boat together1 and no man ever lands on the island4 ,there are other methods that require several more exchanges of places4 #cgg4--,,stealing ,,the ,,castle ,,treasure4 ,here is the best answer1 in eleven manipulations3-- ,treasure down4 ,boy down--treasure up4 ,youth down--boy up4 ,treasure down4 ,man down--youth and treasure up4 ,treasure down4 ,boy down--treasure up4 ,treasure down4 ,youth down--boy up4 ,boy down--treasure up4 ,treasure down4 #cgh4--;,,dominoes ,,in ,,progression4 ,there are twenty-three different ways4 ,you may start with any domino1 except the #d--d and those that bear a #e or #f1 though only certain initial dominoes may be played either way round4 ,if you are given the common difference and the first domino is played1 you have no option as to the other dominoes4 ,therefore all ,i need do is to give the initial domino for all the twenty-three ways1 and state the common difference4 ,this ,i will do as follows3-- ,with a common difference of #a1 the first domino may be either of these3 #j--j1 #j--a1 #a--j1 #j--b1 #a--a1 #b--j1 #j--c1 #a--b1 #b--a1 #c--j1 #j--d1 #a--c1 #b--b1 #c--a1 #a--d1 #b--c1 #c--b1 #b--d1 #c--c1 #c--d4 ,with a difference of #b1 the first domino may be #j--j1 #j--b1 or #j--a4 ,take the last case of all as an example4 ,having played the #j--a1 and the difference being #b1 we are compelled to continue with #a--b1 #b--c1 #c--d4 #d--e1 #e--f4 ,there are three dominoes that can never be used at all4 ,these are #j--e1 #j--f1 and #a--f4 ,if we used a box of dominoes extending to #i--i1 there would be forty different ways4 #cgi4--,,the ,,five ,,dominoes4 ,there are just ten different ways of arranging the dominoes4 ,here is one of them3-- 7#b--j7 7#j--j7 7#j--a7 7#a--d7 7#d--j74 ,i will leave my readers to find the remaining nine for themselves4 #chj4--,,the ,,domino ,,frame ,,puzzle4 ,illustration3 ---------------------------------------- ------------ #b #b #e #e #f #f #f #f #a #a #d - ---------------------------------------- -------- #b #d --- - #b #c - --- #f #c --- ,t ,h ,e - #f #c - --- #c #c --- - #c #a - ,d ,o ,m ,i ,n ,o ,f ,r ,a ,m ,e --- #a --- - #a - --- #e #a --- -,s-,o-,l-,u-,t-,i-,o-,n- - #e #d - --- #c #d --- - #c #f - --- #b #f ---------------------------------------- -------- - #b #a #a #e #e #e #e #d #d #d #d #b #b ---------------------------------------- ------------ ,the illustration is a solution4 ,it will be found that all four sides of the frame add up #dd4 ,the sum of the pips on all the dominoes is #afh1 and if we wish to make the sides sum to #dd1 we must take care that the four corners sum to #h1 because these corners are counted twice1 and #afh added to #h will equal #d times #dd1 which is necessary4 ,there are many different solutions4 ,even in the example given certain interchanges are possible to produce different arrangements4 ,for example1 on the left-hand side the string of dominoes from #b--b down to #c--b may be reversed1 or from #b--f to #c--b1 or from #c--j to #e--c4 ,also1 on the right-hand side we may reverse from #d--c to #a--d4 ,these changes will not affect the correctness of the solution4 #cha4--,,the ,,card ,,frame ,,puzzle4 ,the sum of all the pips on the ten cards is #ee4 ,suppose we are trying to get #ad pips on every side4 ,then #d times #ad is #ef4 ,but each of the four corner cards is added in twice1 so that #ee deducted from #ef1 or #a1 must represent the sum of the four corner cards4 ,this is clearly impossible2 therefore #ad is also impossible4 ,but suppose we came to trying #ah4 ,then #d times #ah is #gb1 and if we deduct #ee we get #ag as the sum of the corners4 ,we need then only try different arrangements with the four corners always summing to #ag1 and we soon discover the following solution3-- ,illustration3 --------------------- #b #aj #f ------------ ------ #c #g --- --- #h #a ------------------- #e #i #d --------------------- ,the final trials are very limited in number1 and must with a little judgment either bring us to a correct solution or satisfy us that a solution is impossible under the conditions we are attempting4 ,the two centre cards on the upright sides can1 of course1 always be interchanged1 but ,i do not call these different solutions4 ,if you reflect in a mirror you get another arrangement1 which also is not considered different4 ,in the answer given1 however1 we may exchange the #e with the #h and the #d with the #a4 ,this is a different solution4 ,there are two solutions with #ah1 four with #ai1 two with #bj1 and two with #bb--ten arrangements in all4 ,readers may like to find all these for themselves4 #chb4--,,the ,,cross ,,of ,,cards4 ,there are eighteen fundamental arrangements1 as follows1 where ,i only give the numbers in the horizontal bar1 since the remainder must naturally fall into their places4 #e #f #a #g #d #b #d #e #f #h #c #e #a #f #h #c #d #e #f #g #c #d #a #g #h #a #d #g #f #h #b #e #a #g #h #b #c #g #f #h #b #e #c #f #h #b #d #g #e #h #a #e #c #g #h #c #d #i #e #f #b #d #c #g #h #b #d #i #e #g #a #d #e #g #h #a #d #i #f #g #b #c #e #g #h #b #c #i #f #g ,it will be noticed that there must always be an odd number in the centre1 that there are four ways each of adding up #bc1 #be1 and #bg1 but only three ways each of summing to #bd and #bf4 #chc4--,,the ,t ,,card ,,puzzle4 ,if we remove the ace1 the remaining cards may he divided into two groups 7each adding up alike7 in four ways2 if we remove #c1 there are three ways2 if #e1 there are four ways2 if #g1 there are three ways2 and if we remove #i1 there are four ways of making two equal groups4 ,there are thus eighteen different ways of grouping1 and if we take any one of these and keep the odd card 7that ,i have called removed7 at the head of the column1 then one set of numbers can be varied in order in twenty-four ways in the column and the other four twenty-four ways in the horizontal1 or together they may be varied in #bd #bd #egf ways4 ,and as there are eighteen such cases1 we multiply this number by #ah and get #aj1cfh1 the correct number of ways of placing the cards4 ,as this number includes the reflections1 we must divide by #b1 but we have also to remember that every horizontal row can change places with a vertical row1 necessitating our multiplying by #b2 so one operation cancels the other4 #chd4--;,,card ,,triangles4 ,the following arrangements of the cards show 7#a7 the smallest possible sum1 #ag2 and 7#b7 the largest possible1 #bc4 #a #g #i #f #d #b #d #h #c #f #c #g #e #b #i #e #a #h ,it will be seen that the two cards in the middle of any side may always be interchanged without affecting the conditions4 ,thus there are eight ways of presenting every fundamental arrangement4 ,the number of fundamentals is eighteen1 as follows3 two summing to #ag1 four summing to #ai1 six summing to #bj1 four summing to #ba1 and two summing to #bc4 ,these eighteen fundamentals1 multiplied by eight 7for the reason stated above71 give #add as the total number of different ways of placing the cards4 #che4--,,strand ,,patience4 ,the reader may find a solution quite easy in a little over #bjj moves1 but1 surprising as it may at first appear1 not more than #fb moves are required4 ,here is the play3 ,by #d ,c up ,i mean a transfer of the #d of clubs with all the cards that rest on it4 #a ,d on space1 #b ,s on space1 #c ,d on space1 #b ,s on #c ,d1 #a ,h on #b ,s1 #b ,c on space1 #a ,d on #b ,c1 #d ,s on space1 #c ,h on #d ,s 7#i moves so far71 #b ,s up on #c ,h 7#c moves71 #e ,h and #e ,d exchanged1 and #d ,c on #e ,d 7#f moves71 #c ,d on #d ,c 7#a71 #f ,s 7with #e ,h7 on space 7#c71 #d ,c up on #e ,h 7#c71 #b ,c up on #c ,d 7#c71 #g ,d on space 7#a71 #f ,c up on #g ,d 7#c71 #h ,s on space 7#a71 #g ,h on #h ,s 7#a71 #h ,c on #i ,d 7#a71 #g ,h on #h ,c 7#a71 #h ,s on #i ,h 7#a71 #g ,h on #h ,s 7#a71 #g ,d up on #h ,c 7#e71 #d ,c up on #e ,d 7#i71 #f ,s up on #g ,h 7#c71 #d ,s up on #e ,h 7#g7 #fb moves in all4 ,this is my record2 perhaps the reader can beat it4 #chf4--;,a ,,trick ,,with ,,dice4 ,all you have to do is to deduct #bej from the result given1 and the three figures in the answer will be the three points thrown with the dice4 ,thus1 in the throw we gave1 the number given would be #chf2 and when we deduct #bej we get #acf1 from which we know that the throws were #a1 #c1 and #f4 ,the process merely consists in giving #ajj;a #aj;b c #bej1 where a1 b1 and c represent the three throws4 ,the result is obvious4 #chg4--,,the ,,village ,,cricket ,,match4 ,illustration3 ,mr4 ,dumkins -- ------------------------ ------------------- ------------------- #a ----------------------- ------------------------ ------------------- ------------------- ----------------------- -- ,mr4 ,podder ,mr4 ,luffey -- ------------------------ ------------------- ----------------------- #b ----------------------- ------------------- ------------------------ -- ,mr4 ,struggles ,the diagram ,no4 #a will show that as neither ,mr4 ,podder nor ,mr4 ,dumkins can ever have been within the crease opposite to that from which he started1 ,mr4 ,dumkins would score nothing by his performance4 ,diagram ,no4 #b will1 however1 make it clear that since ,mr4 ,luffey and ,mr4 ,struggles have1 notwithstanding their energetic but careless movements1 contrived to change places1 the manoeuvre must increase ,mr4 ,struggles's total by one run4 #chh4--,,slow ,,cricket4 ,the captain must have been not out and scored #ba4 ,thus3-- #b men 7each lbw7 #ai #d men 7each caught7 #ag #a man 7run out7 #j #c men 7each bowled7 #i #a man 7captain--not out7 #ba -- -- #aa #ff ,the captain thus scored exactly #ae more than the average of the team4 ,the others who were bowled could only refer to three men1 as the eleventh man would be not out4 ,the reader can discover for himself why the captain must have been that eleventh man4 ,it would not necessarily follow with any figures4 #chi4--,,the ,,football ,,players4 ,the smallest possible number of men is seven4 ,they could be accounted for in three different ways3 #a4 ,two with both arms sound1 one with broken right arm1 and four with both arms broken4 #b4 ,one with both arms sound1 one with broken left arm1 two with broken right arm1 and three with both arms broken4 #c4 ,two with left arm broken1 three with right arm broken1 and two with both arms broken4 ,but if every man was injured1 the last case is the only one that would apply4 #cij4--,,the ,,horse-,,race ,,puzzle4 ,the answer is3 #ab on ,acorn1 #ae on ,bluebottle1 #bj on ,capsule4 #cia4--,,the ,,motor-,,car ,,race4 ,the first point is to appreciate the fact that1 in a race round a circular track1 there are the same number of cars behind one as there are before4 ,all the others are both behind and before4 ,there were thirteen cars in the race1 including ,gogglesmith's car4 ,then one-third of twelve added to three-quarters of twelve will give us thirteen--the correct answer4 #cib4--,,the ,,pebble ,,game4 ,in the case of fifteen pebbles1 the first player wins if he first takes two4 ,then when he holds an odd number and leaves #a1 #h1 or #i he wins1 and when he holds an even number and leaves #d1 #e1 or #ab he also wins4 ,he can always do one or other of these things until the end of the game1 and so defeat his opponent4 ,in the case of thirteen pebbles the first player must lose if his opponent plays correctly4 ,in fact1 the only numbers with which the first player ought to lose are #e and multiples of #h added to #e1 such as #ac1 #ba1 #bi1 etc4 #cic4--,,the ,,two ,,rooks4 ,the second player can always win1 but to ensure his doing so he must always place his rook1 at the start and on every subsequent move1 on the same diagonal as his opponent's rook4 ,he can then force his opponent into a corner and win4 ,supposing the diagram to represent the positions of the rooks at the start1 then1 if ,black played first1 ,white might have placed his rook at ,a and won next move4 ,any square on that diagonal from ,a to ,h will win1 but the best play is always to restrict the moves of the opposing rook as much as possible4 ,if ,white played first1 then ,black should have placed his rook at ,b 7,f would not be so good1 as it gives ,white more scope72 then if ,white goes to ,c1 ,black moves to ,d2 ,white to ,e1 ,black to ,f2 ,white to ,g1 ,black to ,c2 ,white to ,h1 ,black to ,i2 and ,black must win next move4 ,if at any time ,black had failed to move on to the same diagonal as ,white1 then ,white could take ,black's diagonal and win4 r3 black rook ,r3 white rook -------- r -------- ,a -------- -------- -------- ,b,d,f -------- ,r,c,e -------- ,i,g -------- ,h -------- ,,the ,,two ,,rooks4 #cid4--,,puss ,,in ,,the ,,corner4 ,no matter whether he plays first or second1 the player ,a1 who starts the game at #ee1 must win4 ,assuming that ,b adopts the very best lines of play in order to prolong as much as possible his existence1 ,a1 if he has first move1 can always on his #abth move capture ,b2 and if he has the second move1 ,a can always on his #adth move make the capture4 ,his point is always to get diagonally in line with his opponent1 and by going to #cc1 if he has first move1 he prevents ,b getting diagonally in line with himself4 ,here are two good games4 ,the number in front of the hyphen is always ,a's move2 that after the hyphen is ,b's3-- #cc-h1 #cb-ae1 #ca-bb1 #cj-ba1 #bi-ad1 #bb-g1 #ae-f1 #ad-b1 #g-c1 #f-d1 #aa-1 and ,a must capture on his next 7#abth7 move1 -#ac1 #ed-bj1 #ec-bg1 #eb-cd1 #ea-da1 #ej-cd1 #db-bg1 #ce-bj1 #bh-ac1 #ba-f1 #ad-b1 #g-c1 #f-d1 #aa-1 and ,a must capture on his next 7#adth7 move4 #cie4--;,a ,,war ,,puzzle ,,game4 ,the ,britisher can always catch the enemy1 no matter how clever and elusive that astute individual may be2 but curious though it may seem1 the ,british general can only do so after he has paid a somewhat mysterious visit to the particular town marked #a in the map1 going in by #c and leaving by #b1 or entering by #b and leaving by #c4 ,the three towns that are shaded and have no numbers do not really come into the question1 as some may suppose1 for the simple reason that the ,britisher never needs to enter any one of them1 while the enemy cannot be forced to go into them1 and would be clearly ill-advised to do so voluntarily4 ,we may therefore leave these out of consideration altogether4 ,no matter what the enemy may do1 the ,britisher should make the following first nine moves3 ,he should visit towns #bd1 #bj1 #ai1 #ae1 #aa1 #g1 #c1 #a1 #b4 ,if the enemy takes it into his head also to go to town #a1 it will be found that he will have to beat a precipitate retreat the same way that he went in1 or the ,britisher will infallibly catch him in towns #b or #c1 as the case may be4 ,so the enemy will be wise to avoid that north-west corner of the map altogether4 ,illustration ,now1 when the ,british general has made the nine moves that ,i have given1 the enemy will be1 after his own ninth move1 in one of the towns marked #e1 #h1 #aa1 #ac1 #ad1 #af1 #ai1 #ba1 #bd1 or #bg4 ,of course1 if he imprudently goes to #c or #f at this point he will be caught at once4 ,wherever he may happen to be1 the ,britisher goes for him1 and has no longer any difficulty in catching him in eight more moves at most 7seventeen in all7 in one of the following ways4 ,the ,britisher will get to #h when the enemy is at #e1 and win next move2 or he will get to #ai when the enemy is at #bb1 and win next move2 or he will get to #bd when the enemy is at #bg1 and so win next move4 ,it will be found that he can be forced into one or other of these fatal positions4 ,in short1 the strategy really amounts to this3 the ,britisher plays the first nine moves that ,i have given1 and although the enemy does his very best to escape1 our general goes after his antagonist and always driving him away from that north-west corner ultimately closes in with him1 and wins4 ,as ,i have said1 the ,britisher never need make more than seventeen moves in all1 and may win in fewer moves if the enemy plays badly4 ,but after playing those first nine moves it does not matter even if the ,britisher makes a few bad ones4 ,he may lose time1 but cannot lose his advantage so long as he now keeps the enemy from town #a1 and must eventually catch him4 ,this is a complete explanation of the puzzle4 ,it may seem a little complex in print1 but in practice the winning play will now be quite easy to the reader4 ,make those nine moves1 and there ought to be no difficulty whatever in finding the concluding line of play4 ,indeed1 it might almost be said that then it is difficult for the ,british general not to catch the enemy4 ,it is a question of what in chess we call the opposition1 and the visit by the ,britisher to town #a gives him the jump on the enemy1 as the man in the street would say4 ,here is an illustrative example in which the enemy avoids capture as long as it is possible for him to do so4 ,the ,britisher's moves are above the line and the enemy's below it4 ,play them alternately4 #bd #bj #ai #ae #aa #g #c #a #b #f #aj #ad #ah #ai #bj #bd ---------------------------------------- ------- #ac #i #ac #ag #ba #bj #bd #bc #ai #ae #ai #bc #bd #be #bg ,the enemy must now go to #be or ,b1 in either of which towns he is immediately captured4 #cif4--;,a ,,match ,,mystery4 ,if you form the three heaps 7and are therefore the second to draw71 any one of the following thirteen groupings will give you a win if you play correctly3 #ae1 #ad1 #a2 #ae1 #ac1 #b2 #ae1 #ab1 #c2 #ae1 #aa1 #d2 #ae1 #aj1 #e2 #ae1 #i1 #f2 #ae1 #h1 #g2 #ad1 #ac1 #c2 #ad1 #aa1 #e2 #ad1 #i1 #g2 #ac1 #aa1 #f2 #ac1 #aj1 #g2 #ab1 #aa1 #g4 ,the beautiful general solution of this problem is as follows4 ,express the number in every heap in powers of #b1 avoiding repetitions and remembering that #bj #a4 ,then if you so leave the matches to your opponent that there is an even number of every power1 you can win4 ,and if at the start you leave the powers even1 you can always continue to do so throughout the game4 ,take1 as example1 the last grouping given above--#ab1 #aa1 #g4 ,expressed in powers of #b we have-- #ab #h #d - - #aa #h - #b #a #g - #d #b #a ------- #b #b #b #b ------- ,as there are thus two of every power1 you must win4 ,say your opponent takes #g from the #ab heap4 ,he then leaves-- #e - #d - #a #aa #h - #b #a #g - #d #b #a ------- #a #b #b #c ------- ,here the powers are not all even in number1 but by taking #i from the #aa heap you immediately restore your winning position1 thus-- #e - #d - #a #b - - #b - #g - #d #b #a ------- - #b #b #b ------- ,and so on to the end4 ,this solution is quite general1 and applies to any number of matches and any number of heaps4 ,a correspondent informs me that this puzzle game was first propounded by ,mr4 ,w4,m4,f4 ,mellor1 but when or where it was published ,i have not been able to ascertain4 #cig4--,,the ,,montenegrin ,,dice ,,game4 ,the players should select the pairs #e and #i1 and #ac and #ae1 if the chances of winning are to be quite equal4 ,there are #baf different ways in which the three dice may fall4 ,they may add up #e in #f different ways and #i in #be different ways1 making #ca chances out of #baf for the player who selects these numbers4 ,also the dice may add up #ac in #ba different ways1 and #ae in #aj different ways1 thus giving the other player also #ca chances in #baf4 #cih4--,,the ,,cigar ,,puzzle4 ,not a single member of the club mastered this puzzle1 and yet ,i shall show that it is so simple that the merest child can understand its solution--when it is pointed out to him6 ,the large majority of my friends expressed their entire bewilderment4 ,many considered that the theoretical result1 in any case1 is determined by the relationship between the table and the cigars2 others1 regarding it as a problem in the theory of ,probabilities1 arrived at the conclusion that the chances are slightly in favour of the first or second player1 as the case may be4 ,one man took a table and a cigar of particular dimensions1 divided the table into equal sections1 and proceeded to make the two players fill up these sections so that the second player should win4 ,but why should the first player be so accommodating8 ,at any stage he has only to throw down a cigar obliquely across several of these sections entirely to upset ,mr4 #b's calculations6 ,we have to assume that each player plays the best possible2 not that one accommodates the other4 ,the theories of some other friends would be quite sound if the shape of the cigar were that of a torpedo--perfectly symmetrical and pointed at both ends4 ,i will show that the first player should infallibly win1 if he always plays in the best possible manner4 ,examine carefully the following diagram1 ,no4 #a1 and all will be clear4 ,illustration3 #a ,illustration3 #b ,the first player must place his first cigar on end in the exact centre of the table1 as indicated by the little circle4 ,now1 whatever the second player may do throughout1 the first player must always repeat it in an exactly diametrically opposite position4 ,thus1 if the second player places a cigar at ,a1 ,i put one at ,,aa2 he places one at ,b1 ,i put one at ,,bb2 he places one at ,c1 ,i put one at ,,cc2 he places one at ,d1 ,i put one at ,,dd2 he places one at ,e1 ,i put one at ,,ee2 and so on until no more cigars can be placed without touching4 ,as the cigars are supposed to be exactly alike in every respect1 it is perfectly clear that for every move that the second player may choose to make1 it is possible exactly to repeat it on a line drawn through the centre of the table4 ,the second player can always duplicate the first player's move1 no matter where he may place a cigar1 or whether he places it on end or on its side4 ,as the cigars are all alike in every respect1 one will obviously balance over the edge of the table at precisely the same point as another4 ,of course1 as each player is supposed to play in the best possible manner1 it becomes a matter of theory4 ,it is no valid objection to say that in actual practice one would not be sufficiently exact to be sure of winning4 ,if as the first player you did not win1 it would be in consequence of your not having played the best possible4 ,the second diagram will serve to show why the first cigar must be placed on end4 7,and here ,i will say that the first cigar that ,i selected from a box ,i was able so to stand on end1 and ,i am allowed to assume that all the other cigars would do the same47 ,if the first cigar were placed on its side1 as at ,f1 then the second player could place a cigar as at ,g--as near as possible1 but not actually touching ,f4 ,now1 in this position you cannot repeat his play on the opposite side1 because the two ends of the cigar are not alike4 ,it will be seen that ,,gg1 when placed on the opposite side in the same relation to the centre1 intersects1 or lies on top of1 ,f1 whereas the cigars are not allowed to touch4 ,you must therefore put the cigar farther away from the centre1 which would result in your having insufficient room between the centre and the bottom left-hand corner to repeat everything that the other player would do between ,g and the top right-hand corner4 ,therefore the result would not be a certain win for the first player4 #cii4--,,the ,,troublesome ,,eight4 ,illustration3 --------- #d #h #b --------- #c #e #g --------- #g #b #e --------- ,the conditions were to place a different number in each of the nine cells so that the three rows1 three columns1 and two diagonals should each add up #ae4 ,probably the reader at first set himself an impossible task through reading into these conditions something which is not there--a common error in puzzle-solving4 ,if ,i had said a different figure1 instead of a different number1 it would have been quite impossible with the #h placed anywhere but in a corner4 ,and it would have been equally impossible if ,i had said a different whole number4 ,but a number may1 of course1 be fractional1 and therein lies the secret of the puzzle4 ,the arrangement shown in the figure will be found to comply exactly with the conditions3 all the numbers are different1 and the square adds up #ae in all the required eight ways4 #djj4--,,the ,,magic ,,strips4 ,there are of course six different places between the seven figures in which a cut may be made1 and the secret lies in keeping one strip intact and cutting each of the other six in a different place4 ,after the cuts have been made there are a large number of ways in which the thirteen pieces may be placed together so as to form a magic square4 ,here is one of them3-- ,illustration3 ------------- #a #b #c #d #e #f #g ------------ #c #d #e #f #ga #b ----------- #e #f #ga #b #c #d ----------- #ga #b #c #d #e #f ----------- #b #c #d #e #f #ga ----------- #d #e #f #ga #b #c ----------- #f #ga #b #c #d #e ------------ ,the arrangement has some rather interesting features4 ,it will be seen that the uncut strip is at the top1 but it will be found that if the bottom row of figures be placed at the top the numbers will still form a magic square1 and that every successive removal from the bottom to the top 7carrying the uncut strip stage by stage to the bottom7 will produce the same result4 ,if we imagine the numbers to be on seven complete perpendicular strips1 it will be found that these columns could also be moved in succession from left to right or from right to left1 each time producing a magic square4 #dja4--;,,eight ,,jolly ,,gaol-,,birds4 ,there are eight ways of forming the magic square--all merely different aspects of one fundamental arrangement4 ,thus1 if you give our first square a quarter turn you will get the second square2 and as the four sides may be in turn brought to the top1 there are four aspects4 ,these four in turn reflected in a mirror produce the remaining four aspects4 ,now1 of these eight arrangements only four can possibly be reached under the conditions1 and only two of these four can be reached in the fewest possible moves1 which is nineteen4 ,these two arrangements are shown4 ,move the men in the following order3 #e1 #c1 #b1 #e1 #g1 #f1 #d1 #a1 #e1 #g1 #f1 #d1 #a1 #f1 #d1 #h1 #c1 #b1 #g1 and you get the first square4 ,move them thus3 #d1 #a1 #b1 #d1 #a1 #f1 #g1 #a1 #e1 #h1 #a1 #e1 #f1 #g1 #e1 #f1 #d1 #b1 #g1 and you have the arrangement in the second square4 ,in the first case every man has moved1 but in the second case the man numbered #c has never left his cell4 ,therefore ,no4 #c must be the obstinate prisoner1 and the second square must be the required arrangement4 ,illustration3 --------- --------- #e #g #g #d #c - -- -- - - -- -- - #f #d #b #d #h - -- -- - - -- -- - #a #h #c #e #f #a --------- --------- #djb4--,,nine ,,jolly ,,gaol ,,birds4 ,there is a pitfall set for the unwary in this little puzzle4 ,at the start one man is allowed to be placed on the shoulders of another1 so as to give always one empty cell to enable the prisoners to move about without any two ever being in a cell together4 ,the two united prisoners are allowed to add their numbers together1 and are1 of course1 permitted to remain together at the completion of the magic square4 ,but they are obviously not compelled so to remain together1 provided that one of the pair on his final move does not break the condition of entering a cell already occupied4 ,after the acute solver has noticed this point1 it is for him to determine which method is the better one--for the two to be together at the count or to separate4 ,as a matter of fact1 the puzzle can be solved in seventeen moves if the men are to remain together2 but if they separate at the end1 they may actually save a move and perform the feat in sixteen6 ,the trick consists in placing the man in the centre on the back of one of the corner men1 and then working the pair into the centre before their final separation4 ,illustration3 ,a ,b --------- --------- #b #i #d #f #g #b - -- -- - - -- -- - #g #e #c #a #e #i - -- -- - - -- -- - #f #a #h #h #c #d --------- --------- ,here are the moves for getting the men into one or other of the above two positions4 ,the numbers are those of the men in the order in which they move into the cell that is for the time being vacant4 ,the pair is shown in brackets3-- ,place #e on #a4 ,then1 #f1 #i1 #h1 #f1 #d1 7#f71 #b1 #d1 #i1 #c1 #d1 #i1 7#f71 #g1 #f1 #a4 ,place #e on #i4 ,then1 #d1 #a1 #b1 #d1 #f1 7#ad71 #h1 #f1 #a1 #g1 #f1 #a1 7#ad71 #c1 #d1 #i4 ,place #e on #c4 ,then1 #f1 7#h71 #b1 #f1 #d1 #g1 #h1 #d1 #g1 #a1 #f1 #g1 7#h71 #i1 #d1 #c4 ,place #e on #g4 ,then1 #d1 7#ab71 #h1 #d1 #f1 #c1 #b1 #f1 #c1 #i1 #d1 #c1 7#ab71 #a1 #f1 #g4 ,the first and second solutions produce ,diagram ,a2 the second and third produce ,diagram ,b4 ,there are only sixteen moves in every case4 ,having found the fewest moves1 we had to consider how we were to make the burdened man do as little work as possible4 ,it will at once be seen that as the pair have to go into the centre before separating they must take at fewest two moves4 ,the labour of the burdened man can only be reduced by adopting the other method of solution1 which1 however1 forces us to take another move4 #djc4--,,the ,,spanish ,,dungeon4 ,illustration -------------------- -------------------- #a #b #c #d #aj #i #g #d #e #f #g #h #f #e #aa #h #i #aj #aa #ab #a #b #ab #ae #ac #ad #ae #ac #ad #c -------------------- -------------------- ,this can best be solved by working backwards--that is to say1 you must first catch your square1 and then work back to the original position4 ,we must first construct those squares which are found to require the least amount of readjustment of the numbers4 ,many of these we know cannot possibly be reached4 ,when we have before us the most favourable possible arrangements1 it then becomes a question of careful analysis to discover which position can be reached in the fewest moves4 ,i am afraid1 however1 it is only after considerable study and experience that the solver is able to get such a grasp of the various areas of disturbance and methods of circulation that his judgment is of much value to him4 ,the second diagram is a most favourable magic square position4 ,it will be seen that prisoners #d1 #h1 #ac1 and #ad are left in their original cells4 ,this position may be reached in as few as thirty-seven moves4 ,here are the moves3 #ae1 #ad1 #aj1 #f1 #g1 #c1 #b1 #g1 #f1 #aa1 #c1 #b1 #g1 #f1 #aa1 #aj1 #ad1 #c1 #b1 #aa1 #aj1 #i1 #e1 #a1 #f1 #aj1 #i1 #e1 #a1 #f1 #aj1 #i1 #e1 #b1 #ab1 #ae1 #c4 ,this short solution will probably surprise many readers who may not find a way under from sixty to a hundred moves4 ,the clever prisoner was ,no4 #f1 who in the original illustration will be seen with his arms extended calling out the moves4 ,he and ,no4 #aj did most of the work1 each changing his cell five times4 ,no4 #ab1 the man with the crooked leg1 was lame1 and therefore fortunately had only to pass from his cell into the next one when his time came round4 #djd4--,,the ,,siberian ,,dungeons4 ,illustration -------------------- #h #e #aj #aa #af #ac #b #c #a #ab #g #ad #i #d #ae #f -------------------- ,in attempting to solve this puzzle it is clearly necessary to seek such magic squares as seem the most favourable for our purpose1 and then carefully examine and try them for fewest moves4 ,of course it at once occurs to us that if we can adopt a square in which a certain number of men need not leave their original cells1 we may save moves on the one hand1 but we may obstruct our movements on the other4 ,for example1 a magic square may be formed with the #f1 #g1 #ac1 and #af unmoved2 but in such case it is obvious that a solution is impossible1 since cells #ad and #ae can neither be left nor entered without breaking the condition of no two men ever being in the same cell together4 ,the following solution in fourteen moves was found by ,mr4 ,g4 ,wotherspoon3 #h-ag1 #af-ba1 #f-af1 #ad-h1 #e-ah1 #d-ad1 #c-bd1 #aa-bj1 #aj-ai1 #b-bc1 #ac-bb1 #ab-f1 #a-e1 #i-ac4 ,as this solution is in what ,i consider the theoretical minimum number of moves1 ,i am confident that it cannot be improved upon1 and on this point ,mr4 ,wotherspoon is of the same opinion4 #dje4--;,,card ,,magic ,,squares4 ,arrange the cards as follows for the three new squares3-- #c #b #d #f #e #g #i #h #aj #d #c #b #g #f #e #aj #i #h #b #d #c #e #g #f #h #aj #i ,three aces and one ten are not used4 ,the summations of the four squares are thus3 #i1 #ae1 #ah1 and #bg--;all different1 as required4 #djf4--,,the ,,eighteen ,,dominoes4 ,illustration ,the illustration explains itself4 ,it will be found that the pips in every column1 row1 and long diagonal add up #ah1 as required4 #djg4--,,two ,,new ,,magic ,,squares4 ,here are two solutions that fulfil the conditions3-- ,illustration3 ,,subtracting ,,dividing #aa #d #ad #ac #cf #h #ed #bg #af #g #a #b #baf #ab #a #b #f #e #c #ab #f #c #d #gb #i #ai #h #ae #i #ah #bd #ajh ,the first1 by subtracting1 has a constant #h1 and the associated pairs all have a difference of #d4 ,the second square1 by dividing1 has a constant #i1 and all the associated pairs produce #c by division4 ,these are two remarkable and instructive squares4 #djh4--,,magic ,,squares ,,of ,,two ,,degrees4 ,the following is the square that ,i constructed4 ,as it stands the constant is #bfj4 ,if for every number you substitute1 in its allotted place1 its square1 then the constant will be #aa1ahj4 ,readers can write out for themselves the second degree square4 ,illustration3 #g #ec #da #bg #b #eb #dh #cj #ab #eh #ch #bd #ac #fc #ce #ag -------------------------- #ea #a #bi #dg #ed #h #bh #db #fd #ad #ah #cf #eg #aa #bc #cg -------------------------- #be #dc #ee #e #cb #df #ej #d #bb #dj #fj #aj #ai #cc #fa #ae -------------------------- #de #ca #c #di #dd #bf #f #ef #cd #bj #af #fb #ci #ba #i #ei ,the main key to the solution is the pretty law that if eight numbers sum to #bfj and their squares to #aa1ahj1 then the same will happen in the case of the eight numbers that are complementary to #fe4 ,thus #a #ah #bc #bf #ca #dh #ef #eg #bfj1 and the sum of their squares is #aa1ahj4 ,therefore #fd #dg #db #ci #cd #ag #i #h 7obtained by subtracting each of the above numbers from #fe7 will sum to #bfj and their squares to #aa1ahj4 ,note that in every one of the sixteen smaller squares the two diagonals sum to #fe4 ,there are four columns and four rows with their complementary columns and rows4 ,let us pick out the numbers found in the #bnd1 #ast1 #dth1 and #crd rows and arrange them thus 3-- ,illustration3 #a #h #bh #bi #db #dg #ea #ed #b #g #bg #cj #da #dh #eb #ec #c #f #bf #ca #dd #de #di #ef #d #e #be #cb #dc #df #ej #ee ,here each column contains four consecutive numbers cyclically arranged1 four running in one direction and four in the other4 ,the numbers in the #bnd1 #eth1 #crd1 and #hth columns of the square may be similarly grouped4 ,the great difficulty lies in discovering the conditions governing these groups of numbers1 the pairing of the complementaries in the squares of four and the formation of the diagonals4 ,but when a correct solution is shown1 as above1 it discloses all the more important keys to the mystery4 ,i am inclined to think this square of two degrees the most elegant thing that exists in magics4 ,i believe such a magic square cannot be constructed in the case of any order lower than #h4 #dji4--,,the ,,baskets ,,of ,,plums4 ,as the merchant told his man to distribute the contents of one of the baskets of plums among some children1 it would not be permissible to give the complete basketful to one child2 and as it was also directed that the man was to give plums to every child1 so that each should receive an equal number1 it would also not be allowed to select just as many children as there were plums in a basket and give each child a single plum4 ,consequently1 if the number of plums in every basket was a prime number1 then the man would be correct in saying that the proposed distribution was quite impossible4 ,our puzzle1 therefore1 resolves itself into forming a magic square with nine different prime numbers4 ,illustration ,a ,b --------------- --------------- #g #fa #dc #hc #bi #aja #gc #cg #a #hi #ga #ec #ca #ac #fg #da #aac #ei --------------- --------------- ,c ,d --------------- --------------- #ajc #gi #cg #affi #aii #abdi #g #gc #aci #fai #ajci #adei #aji #fg #dc #hbi #ahgi #dji --------------- --------------- ,in ,diagram ,a we have a magic square in prime numbers1 and it is the one giving the smallest constant sum that is possible4 ,as to the little trap ,i mentioned1 it is clear that ,diagram ,a is barred out by the words every basket contained plums1 for one plum is not plums4 ,and as we were referred to the baskets1 as shown in the illustration1 it is perfectly evident1 without actually attempting to count the plums1 that there are at any rate more than #g plums in every basket4 ,therefore ,c is also1 strictly speaking1 barred4 ,numbers over #bj and under1 say1 #bej would certainly come well within the range of possibility1 and a large number of arrangements would come within these limits4 ,diagram ,b is one of them4 ,of course we can allow for the false bottoms that are so frequently used in the baskets of fruitsellers to make the basket appear to contain more fruit than it really does4 ,several correspondents assumed 7on what grounds ,i cannot think7 that in the case of this problem the numbers cannot be in consecutive arithmetical progression1 so ,i give ,diagram ,d to show that they were mistaken4 ,the numbers are #aii1 #dji1 #fai1 #hbi1 #a1jci1 #a1bdi1 #a1dei1 #a1ffi1 and #a1hgi--;all primes with a common difference of #baj4 #daj4--,,the ,,mandarin',s ,t ,,puzzle4 ,there are many different ways of arranging the numbers1 and either the #b or the #c may be omitted from the ,t enclosure4 ,the arrangement that ,i give is a nasik square4 ,out of the total of #bh1hjj nasik squares of the fifth order this is the only one 7with its one reflection7 that fulfils the ,t condition4 ,this puzzle was suggested to me by ,dr4 ,c4 ,planck4 ,illustration3 ,,the ,,mandarin',s ,t ,,puzzle4 ------------------------- #ai #bc #aa #e #g #a #aj #ag #bd #ac #bb #ad #c #f #bj #h #af #be #ab #d #ae #b #i #ah #ba ------------------------- #daa4--;,a ,,magic ,,square ,,of ,,composites4 ,the problem really amounts to finding the smallest prime such that the next higher prime shall exceed it by #aj at least4 ,if we write out a little list of primes1 we shall not need to exceed #aej to discover what we require1 for after #aac the next prime is #abg4 ,we can then form the square in the diagram1 where every number is composite4 ,this is the solution in the smallest numbers4 ,we thus see that the answer is arrived at quite easily1 in a square of the third order1 by trial4 ,but ,i propose to show how we may get an answer 7not1 it is true1 the one in smallest numbers7 without any tables or trials1 but in a very direct and rapid manner4 ,illustration --------------- #aba #aad #aai #aaf #aah #abj #aag #abb #aae --------------- ,first write down any consecutive numbers1 the smallest being greater than #a--say1 #b1 #c1 #d1 #e1 #f1 #g1 #h1 #i1 #aj4 ,the only factors in these numbers are #b1 #c1 #e1 and #g4 ,we therefore multiply these four numbers together and add the product1 #baj1 to each of the nine numbers4 ,the result is the nine consecutive composite numbers1 #bab to #bbj inclusive1 with which we can form the required square4 ,every number will necessarily be divisible by its difference from #baj4 ,it will be very obvious that by this method we may find as many consecutive composites as ever we please4 ,suppose1 for example1 we wish to form a magic square of sixteen such numbers2 then the numbers #b to #ag contain the factors #b1 #c1 #e1 #g1 #aa1 #ac1 and #ag1 which1 multiplied together1 make #eajeaj to be added to produce the sixteen numbers #eajeab to #eajebg inclusive1 all of which are composite as before4 ,but1 as ,i have said1 these are not the answers in the smallest numbers3 for if we add #ebc to the numbers #a to #af1 we get sixteen consecutive composites2 and if we add #a1cbg to the numbers #a to #be1 we get twenty-five consecutive composites1 in each case the smallest numbers possible4 ,yet if we required to form a magic square of a hundred such numbers1 we should find it a big task by means of tables1 though by the process ,i have shown it is quite a simple matter4 ,even to find thirty-six such numbers you will search the tables up to #aj1jjj without success1 and the difficulty increases in an accelerating ratio with each square of a larger order4 #dab4--,,the ,,magic ,,knight',s ,,tour4 -------------------------------- #df #ee #dd #ai #eh #i #bb #g -------------------------------- #dc #ah #dg #ef #ba #f #ei #aj -------------------------------- #ed #de #bj #da #ab #eg #h #bc -------------------------------- #ag #db #ec #dh #e #bd #aa #fj -------------------------------- #eb #c #cb #ac #dj #fa #cd #be -------------------------------- #ca #af #di #d #cc #bh #cg #fb -------------------------------- #b #ea #ad #bi #fd #ci #bf #ce -------------------------------- #ae #cj #a #ej #bg #cf #fc #ch -------------------------------- ,here each successive number 7in numerical order7 is a knight's move from the preceding number1 and as #fd is a knight's move from #a1 the tour is re-entrant4 ,all the columns and rows add up #bfj4 ,unfortunately1 it is not a perfect magic square1 because the diagonals are incorrect1 one adding up #bfd and the other #bef--requiring only the transfer of #d from one diagonal to the other4 ,i think this is the best result that has ever been obtained 7either re-entrant or not71 and nobody can yet say whether a perfect solution is possible or impossible4 #dac4--;,a ,,chessboard ,,fallacy4 ,illustration ,the explanation of this little fallacy is as follows4 ,the error lies in assuming that the little triangular piece1 marked ,c1 is exactly the same height as one of the little squares of the board4 ,as a matter of fact1 its height 7if we make the sixty-four squares each a square inch7 will be #aa_/g in4 ,consequently the rectangle is really #ia_/g in4 by #g in41 so that the area is sixty-four square inches in either case4 ,now1 although the pieces do fit together exactly to form the perfect rectangle1 yet the directions of the horizontal lines in the pieces will not coincide4 ,the new diagram above will make everything quite clear to the reader4 #dad4--,,who ,,was ,,first8 ,biggs1 who saw the smoke1 would be first2 ,carpenter1 who saw the bullet strike the water1 would be second2 and ,anderson1 who heard the report1 would be last of all4 #dae4--;,a ,,wonderful ,,village4 ,when the sun is in the horizon of any place 7whether in ,japan or elsewhere71 he is the length of half the earth's diameter more distant from that place than in his meridian at noon4 ,as the earth's semi-diameter is nearly #d1jjj miles1 the sun must be considerably more than #c1jjj miles nearer at noon than at his rising1 there being no valley even the hundredth part of #a1jjj miles deep4 #daf4--;,a ,,calendar ,,puzzle4 ,the first day of a century can never fall on a ,sunday2 nor on a ,wednesday or a ,friday4 #dag4--,,the ,,tiring-,,irons4 ,i will give my complete working of the solution1 so that readers may see how easy it is when you know how to proceed4 ,and first of all1 as there is an even number of rings1 ,i will say that they may all be taken off in one-third of 7#b7n #a7 - #b7 moves2 and since n in our case is #ad1 all the rings may be taken off in #aj1ibb moves4 ,then ,i say #aj1ibb - #i1iii #ibc1 and proceed to find the position when only #ibc out of the #aj1ibb moves remain to be made4 ,here is the curious method of doing this4 ,it is based on the binary scale method used by ,monsieur ,l4 ,gros1 for an account of which see ,w4,w4 ,rouse ,ball's ,mathematical ,recreations4 ,divide #ibc by #b1 and we get #dfa and the remainder #a2 divide #dfa by #b1 and we get #bcj and the remainder #a2 divide #bcj by #b1 and we get #aae and the remainder nought4 ,keep on dividing by #b in this way as long as possible1 and all the remainders will be found to be #a1 #a1 #a1 #j1 #j1 #a1 #a1 #j1 #a1 #a1 the last remainder being to the left and the first remainder to the right4 ,as there are fourteen rings and only ten figures1 we place the difference1 in the form of four noughts1 in brackets to the left1 and bracket all those figures that repeat a figure on their left4 ,then we get the following arrangement3 7#j #j #j #j7 #a 7#a #a7 #j 7#j7 #a 7#a7 #j #a 7#a74 ,this is the correct answer to the puzzle1 for if we now place rings below the line to represent the figures in brackets and rings on the line for the other figures1 we get the solution in the required form1 as below3-- ,o ,o ,o ,,oo ------------------------- ,,oooo ,,oo ,o ,o ,o ,this is the exact position of the rings after the #i1iiith move has been made1 and the reader will find that the method shown will solve any similar question1 no matter how many rings are on the tiring-irons4 ,but in working the inverse process1 where you are required to ascertain the number of moves necessary in order to reach a given position of the rings1 the rule will require a little modification1 because it does not necessarily follow that the position is one that is actually reached in course of taking off all the rings on the irons1 as the reader will presently see4 ,i will here state that where the total number of rings is odd the number of moves required to take them all off is one-third of 7#b7n #a7 - #a74 ,with n rings 7where n is odd7 there are #bn positions counting all on and all off4 ,in 7#a_/c77b7n #a7 #b7 positions they are all removed4 ,the number of positions not used is 7#a_/c77bn - #b74 ,with n rings 7where n is even7 there are #bn positions counting all on and all off4 ,in 7#b7n #a7 #a7 positions they are all removed4 ,the number of positions not used is here 7#a_/c77bn - #a74 ,it will be convenient to tabulate a few cases4 ---------------------------------------- -- ,no4 of ,total ,positions ,positions ,rings4 ,positions4 used4 not used4 ---------------------------------------- -- #a #b #b #j #c #h #f #b #e #cb #bb #aj #g #abh #hf #db #i #eab #cdb #agj #b #d #c #a #d #af #aa #e #f #fd #dc #ba #h #bef #aga #he #aj #ajbd #fhc #cda ---------------------------------------- -- ,note first that the number of positions used is one more than the number of moves required to take all the rings off1 because we are including all on which is a position but not a move4 ,then note that the number of positions not used is the same as the number of moves used to take off a set that has one ring fewer4 ,for example1 it takes #he moves to remove #g rings1 and the #db positions not used are exactly the number of moves required to take off a set of #f rings4 ,the fact is that if there are #g rings and you take off the first #f1 and then wish to remove the #gth ring1 there is no course open to you but to reverse all those #db moves that never ought to have been made4 ,in other words1 you must replace all the #g rings on the loop and start afresh6 ,you ought first to have taken off #e rings1 to do which you should have taken off #c rings1 and previously to that #a ring4 ,to take off #f you first remove #b and then #d rings4 #dah4--,,such ,a ,,getting ,,upstairs4 ,number the treads in regular order upwards1 #a to #h4 ,then proceed as follows3 #a 7step back to floor71 #a1 #b1 #c 7#b71 #c1 #d1 #e 7#d71 #e1 #f1 #g 7#f71 #g1 #h1 landing 7#h71 landing4 ,the steps in brackets are taken in a backward direction4 ,it will thus be seen that by returning to the floor after the first step1 and then always going three steps forward for one step backward1 we perform the required feat in nineteen steps4 #dai4--,,the ,,five ,,pennies4 ,illustration ,first lay three of the pennies in the way shown in ,fig4 #a4 ,now hold the remaining two pennies in the position shown in ,fig4 #b1 so that they touch one another at the top1 and at the base are in contact with the three horizontally placed coins4 ,then the five pennies will be equidistant1 for every penny will touch every other penny4 #dbj4--,,the ,,industrious ,,bookworm4 ,the hasty reader will assume that the bookworm1 in boring from the first to the last page of a book in three volumes1 standing in their proper order on the shelves1 has to go through all three volumes and four covers4 ,this1 in our case1 would mean a distance of #i in41 which is a long way from the correct answer4 ,you will find1 on examining any three consecutive volumes on your shelves1 that the first page of ,vol4 ,i4 and the last page of ,vol4 ,,iii4 are actually the pages that are nearest to ,vol4 ,,ii41 so that the worm would only have to penetrate four covers 7together1 in47 and the leaves in the second volume 7#c in471 or a distance of #c inches1 in order to tunnel from the first page to the last4 #dba4--;,a ,,chain ,,puzzle4 ,to open and rejoin a link costs threepence4 ,therefore to join the nine pieces into an endless chain would cost #bs4 #c;d41 whereas a new chain would cost #bs4 #b;d4 ,but if we break up the piece of eight links1 these eight will join together the remaining eight pieces at a cost of #bs4 ,but there is a subtle way of even improving on this4 ,break up the two pieces containing three and four links respectively1 and these seven will join together the remaining seven pieces at a cost of only #as4 #i;d4 #dbb4--,,the ,,sabbath ,,puzzle4 ,the way the author of the old poser proposed to solve the difficulty was as follows3 ,from the ,jew's abode let the ,christian and the ,turk set out on a tour round the globe1 the ,christian going due east and the ,turk due west4 ,readers of ,edgar ,allan ,poe's story1 ,three ,sundays in a ,week1 or of ,jules ,verne's ,round the ,world in ,eighty ,days1 will know that such a proceeding will result in the ,christian's gaining a day and in the ,turk's losing a day1 so that when they meet again at the house of the ,jew their reckoning will agree with his1 and all three may keep their ,sabbath on the same day4 ,the correctness of this answer1 of course1 depends on the popular notion as to the definition of a day--the average duration between successive sun-rises4 ,it is an old quibble1 and quite sound enough for puzzle purposes4 ,strictly speaking1 the two travellers ought to change their reckonings on passing the #ahjth meridian2 otherwise we have to admit that at the ,north or ,south ,pole there would only be one ,sabbath in seven years4 #dbc4--,,the ,,ruby ,,brooch4 ,in this case we were shown a sketch of the brooch exactly as it appeared after the four rubies had been stolen from it4 ,the reader was asked to show the positions from which the stones may have been taken2 for it is not possible to show precisely how the gems were originally placed1 because there are many such ways4 ,but an important point was the statement by ,lady ,littlewood's brother3 ,i know the brooch well4 ,it originally contained forty-five stones1 and there are now only forty-one4 ,somebody has stolen four rubies1 and then reset as small a number as possible in such a way that there shall always be eight stones in any of the directions you have mentioned4 ,illustration ,the diagram shows the arrangement before the robbery4 ,it will be seen that it was only necessary to reset one ruby--the one in the centre4 ,any solution involving the resetting of more than one stone is not in accordance with the brother's statement1 and must therefore be wrong4 ,the original arrangement was1 of course1 a little unsymmetrical1 and for this reason the brooch was described as rather eccentric4 #dbd4--,,the ,,dovetailed ,,block4 ,illustration ,the mystery is made clear by the illustration4 ,it will be seen at once how the two pieces slide together in a diagonal direction4 #dbe4--;,,jack ,,and ,,the ,,beanstalk4 ,the serious blunder that the artist made in this drawing was in depicting the tendrils of ,illustration the bean climbing spirally as at ,a above1 whereas the ,french bean1 or scarlet runner1 the variety clearly selected by the artist in the absence of any authoritative information on the point1 always climbs as shown at ,b4 ,very few seem to be aware of this curious little fact4 ,though the bean always insists on a sinistrorsal growth1 as ,b1 the hop prefers to climb in a dextrorsal manner1 as ,a4 ,why1 is one of the mysteries that ,nature has not yet unfolded4 #dbf4--,,the ,,hymn-,,board ,,poser4 ,this puzzle is not nearly so easy as it looks at first sight4 ,it was required to find the smallest possible number of plates that would be necessary to form a set for three hymn-boards1 each of which would show the five hymns sung at any particular service1 and then to discover the lowest possible cost for the same4 ,the hymn-book contains #gjj hymns1 and therefore no higher number than #gjj could possibly be needed4 ,now1 as we are required to use every legitimate and practical method of economy1 it should at once occur to us that the plates must be painted on both sides2 indeed1 this is such a common practice in cases of this kind that it would readily occur to most solvers4 ,we should also remember that some of the figures may possibly be reversed to form other figures2 but as we were given a sketch of the actual shapes of these figures when painted on the plates1 it would be seen that though the #f's may be turned upside down to make #i's1 none of the other figures can be so treated4 ,it will be found that in the case of the figures #a1 #b1 #c1 #d1 and #e1 thirty-three of each will be required in order to provide for every possible emergency2 in the case of #g1 #h1 and #j1 we can only need thirty of each2 while in the case of the figure #f 7which may be reversed for the figure #i7 it is necessary to provide exactly forty-two4 ,it is therefore clear that the total number of figures necessary is #big2 but as the figures are painted on both sides of the plates1 only #adi such plates are required4 ,at first it would appear as if one of the plates need only have a number on one side1 the other side being left blank4 ,but here we come to a rather subtle point in the problem4 ,readers may have remarked that in real life it is sometimes cheaper when making a purchase to buy more articles than we require1 on the principle of a reduction on taking a quantity3 we get more articles and we pay less4 ,thus1 if we want to buy ten apples1 and the price asked is a penny each if bought singly1 or ninepence a dozen1 we should both save a penny and get two apples more than we wanted by buying the full twelve4 ,in the same way1 since there is a regular scale of reduction for plates painted alike1 we actually save by having two figures painted on that odd plate4 ,supposing1 for example1 that we have thirty plates painted alike with #e on one side and #f on the other4 ,the rate would be #d;d41 and the cost #aas4 #aj;d4 ,but if the odd plate with1 say1 only a #e on one side of it have a #f painted on the other side1 we get thirty-one plates at the reduced rate of #d;d41 thus saving a farthing on each of the previous thirty1 and reducing the cost of the last one from #as4 to #d;d4 ,but even after these points are all seen there comes in a new difficulty3 for although it will be found that all the #h's may be on the backs of the #g's1 we cannot have all the #b's on the backs of the #a's1 nor all the #d on the backs of the #c's1 etc4 ,there is a great danger1 in our attempts to get as many as possible painted alike1 of our so adjusting the figures that some particular combination of hymns cannot be represented4 ,here is the solution of the difficulty that was sent to the vicar of ,chumpley ,st4 ,winifred4 ,where the sign ,x is placed between two figures1 it implies that one of these figures is on one side of the plate and the other on the other side4 d4 s4 d4 #ca plates painted #e ,x #i #d #j #aa #g #cj #g ,x #h #d #j #aa #aj #ba #a ,x #b #g #j #ab #c #ba #c ,x #j #g #j #ab #c #ab #a ,x #c #i #j #i #c #ab #b ,x #d #i #j #i #c #ab #i ,x #d #i #j #i #c #h #d ,x #j #aj #j #f #aj #a #e ,x #d #ab #j #a #j #a #e ,x #j #ab #j #a #j #adi plates #f;d4 each #c #ad #f ---------- #g #ai #a ,of course1 if we could increase the number of plates1 we might get the painting done for nothing1 but such a contingency is prevented by the condition that the fewest possible plates must be provided4 ,this puzzle appeared in ,tit-,bits1 and the following remarks1 made by me in the issue for #aath ,december #ahig1 may be of interest4 ,the ,hymn-,board ,poser seems to have created extraordinary interest4 ,the immense number of attempts at its solution sent to me from all parts of the ,united ,kingdom and from several ,continental countries show a very kind disposition amongst our readers to help the worthy vicar of ,chumpley ,st4 ,winifred over his parochial difficulty4 ,every conceivable estimate1 from a few shillings up to as high a sum as #a1cdg1 #ajs41 seems to have come to hand4 ,but the astonishing part of it is that1 after going carefully through the tremendous pile of correspondence1 ,i find that only one competitor has succeeded in maintaining the reputation of the ,tit-,bits solvers for their capacity to solve anything1 and his solution is substantially the same as the one given above1 the cost being identical4 ,some of his figures are differently combined1 but his grouping of the plates1 as shown in the first column1 is exactly the same4 ,though a large majority of competitors clearly hit upon all the essential points of the puzzle1 they completely collapsed in the actual arrangement of the figures4 ,according to their methods1 some possible selection of hymns1 such as #aaa1 #aab1 #aba1 #abb1baa1 cannot be set up4 ,a few correspondents suggested that it might be possible so to paint the #g's that upside down they would appear as #b's or #d's2 but this would1 of course1 be barred out by the fact that a representation of the actual figures to be used was given4 #dbg4--,,pheasant-,,shooting4 ,the arithmetic of this puzzle is very easy indeed4 ,there were clearly #bd pheasants at the start4 ,of these #af were shot dead1 #a was wounded in the wing1 and #g got away4 ,the reader may have concluded that the answer is1 therefore1 that seven remained4 ,but as they flew away it is clearly absurd to say that they remained4 ,had they done so they would certainly have been killed4 ,must we then conclude that the #ag that were shot remained1 because the others flew away8 ,no2 because the question was not how many remained8 but how many still remained8 ,now the poor bird that was wounded in the wing1 though unable to fly1 was very active in its painful struggles to run away4 ,the answer is1 therefore1 that the #af birds that were shot dead still remained1 or remained still4 #dbh4--,,the ,,gardener ,,and ,,the ,,cook4 ,nobody succeeded in solving the puzzle1 so ,i had to let the cat out of the bag--an operation that was dimly foreshadowed by the puss in the original illustration4 ,but ,i first reminded the reader that this puzzle appeared on ,april #a1 a day on which none of us ever resents being made an ,april ,fool2 though1 as ,i practically gave the thing away by specially drawing attention to the fact that it was ,all ,fools' ,day1 it was quite remarkable that my correspondents1 without a single exception1 fell into the trap4 ,one large body of correspondents held that what the cook loses in stride is exactly made up in greater speed2 consequently both advance at the same rate1 and the result must be a tie4 ,but another considerable section saw that1 though this might be so in a race #bjj ft4 straight away1 it could not really be1 because they each go a stated distance at every bound1 and as #ajj is not an exact multiple of #c1 the gardener at his thirty-fourth bound will go #b ft4 beyond the mark4 ,the gardener will1 therefore1 run to a point #ajb ft4 straight away and return 7#bjd ft4 in all71 and so lose by #d ft4 ,this point certainly comes into the puzzle4 ,but the most important fact of all is this1 that it so happens that the gardener was a pupil from the ,horticultural ,college for ,lady ,gardeners at1 if ,i remember aright1 ,swanley2 while the cook was a very accomplished ,french chef of the hemale persuasion6 ,therefore she 7the gardener7 made three bounds to his 7the cook's7 two4 ,it will now be found that while the gardener is running her #bjd ft4 in #fh bounds of #c ft41 the somewhat infirm old cook can only make #dea_/c of his #b ft4 bounds1 which equals #ij ft4 #h in4 ,the result is that the lady gardener wins the race by #aji ft4 #d in4 at a moment when the cook is in the air1 one-third through his #dfth bound4 ,the moral of this puzzle is twofold3 7#a7 ,never take things for granted in attempting to solve puzzles2 7#b7 always remember ,all ,fools' ,day when it comes round4 ,i was not writing of any gardener and cook1 but of a particular couple1 in a race that ,i witnessed4 ,the statement of the eye-witness must therefore be accepted3 as the reader was not there1 he cannot contradict it4 ,of course the information supplied was insufficient1 but the correct reply was3 ,assuming the gardener to be the 'he1' the cook wins by #d ft42 but if the gardener is the 'she1' then the gardener wins by #aji ft4 #d in4 ,this would have won the prize4 ,curiously enough1 one solitary competitor got on to the right track1 but failed to follow it up4 ,he said3 ,is this a regular ,april #a catch1 meaning that they only ran #f ft4 each1 and consequently the race was unfinished8 ,if not1 ,i think the following must be the solution1 supposing the gardener to be the 'he' and the cook the 'she4' ,though his solution was wrong even in the case he supposed1 yet he was the only person who suspected the question of sex4 #dbi4--,,placing ,,halfpennies4 ,thirteen coins may be placed as shown on page #beb4 #dcj4--;,,find ,,the ,,man',s ,,wife4 ,there is no guessing required in this puzzle4 ,it is all a question of elimination4 ,if we can pair off any five of the ladies with their respective husbands1 other than husband ,no4 #aj1 then the remaining lady must be ,no4 #aj's wife4 ,illustration3 ,,placing ,,halfpennies4 ,i will show how this may be done4 ,no4 #h is seen carrying a lady's parasol in the same hand with his walking-stick4 ,but every lady is provided with a parasol1 except ,no4 #c2 therefore ,no4 #c may be safely said to be the wife of ,no4 #h4 ,then ,no4 #ab is holding a bicycle1 and the dress-guard and make disclose the fact that it is a lady's bicycle4 ,the only lady in a cycling skirt is ,no4 #e2 therefore we conclude that ,no4 #e is ,no4 #ab's wife4 ,next1 the man ,no4 #f has a dog1 and lady ,no4 #aa is seen carrying a dog chain4 ,so we may safely pair ,no4 #f with ,no4 #aa4 ,then we see that man ,no4 #b is paying a newsboy for a paper4 ,but we do not pay for newspapers in this way before receiving them1 and the gentleman has apparently not taken one from the boy4 ,but lady ,no4 #i is seen reading a paper4 ,the inference is obvious--that she has sent the boy to her husband for a penny4 ,we therefore pair ,no4 #b with ,no4 #i4 ,we have now disposed of all the ladies except ,nos4 #a and #g1 and of all the men except ,nos4 #d and #aj4 ,on looking at ,no4 #d we find that he is carrying a coat over his arm1 and that the buttons are on the left side2--not on the right1 as a man wears them4 ,so it is a lady's coat4 ,but the coat clearly does not belong to ,no4 #a1 as she is seen to be wearing a coat already1 while ,no4 #g lady is very lightly clad4 ,we therefore pair ,no4 #g lady with man ,no4 #d4 ,now the only lady left is ,no4 #a1 and we are consequently forced to the conclusion that she is the wife of ,no4 #aj4 ,this is therefore the correct answer4 ,,index4 ,abbot's ,puzzle1 ,the1 #bj1 #afa4 ---- ,window1 ,the1 #hg1 #bac4 ,academic ,courtesies1 #ah1 #afj4 ,acrostic ,puzzle1 ,an1 #hd1 #baj4 ,adam and ,eve and the ,apples1 #ah4 ,aeroplanes1 ,the ,two1 #b1 #adh4 ,age and ,kinship ,puzzles1 #f4 ---- ,concerning ,tommy's1 #g1 #aec4 ---- ,mamma's1 #g1 #aeb4 ---- ,mrs4 ,timpkins's1 #g1 #aeb4 ---- ,rover's1 #g1 #aeb4 ,ages1 ,the ,family1 #g1 #aeb4 ---- ,their1 #g1 #aeb4 ,alcuin1 ,abbot1 #bj1 #aab4 ,almonds1 ,the ,nine1 #fd1 #aie4 ,amazons1 ,the1 #id1 #bba4 ,andrews1 ,w4,s41 #abe4 ,apples1 ,a ,deal in1 #c1 #adi4 ---- ,buying1 #f1 #aea4 ---- ,the ,ten1 #fd1 #aie4 ,approximations in ,dissection1 #bh4 ,arithmetical and ,algebraical ,problems1 #a4 ---- ,various1 #ag4 ,arthur's ,knights1 ,king1 #gg1 #bjc4 ,artillerymen's ,dilemma1 #bf1 #afg4 ,asparagus1 ,bundles of1 #adj4 ,aspects all due ,south1 #acg4 ,associated ,magic ,squares1 #abj4 ,axiom1 ,a ,puzzling1 #ach4 ,bachet de ,mziriac1 #ij1 #aji1 #aab4 ,bachet's ,square1 #ij1 #baf4 ,ball ,problem1 ,the1 #ea1 #ahc4 ,ball1 ,w4,w4 ,rouse1 #aji1 #bjd1 #bdh4 ,balls1 ,the ,glass1 #gh1 #bjd4 ,banker's ,puzzle1 ,the1 #be1 #afe4 ,bank ,holiday ,puzzle1 ,a1 #gc1 #bja4 ,banner ,puzzle1 ,the1 #df1 #agi4 ---- ,st4 ,george's1 #ej1 #ahb4 ,barrel ,puzzle1 ,the1 #aji1 #bce4 ,barrels of ,balsam1 ,the1 #hb1 #bjh4 ,beanfeast ,puzzle1 ,a1 #b1 #adh4 ,beef and ,sausages1 #c1 #adi4 ,beer1 ,the ,barrel of1 #ac1 #aee4 ,bell-ropes1 ,stealing the1 #di1 #aha4 ,bells1 ,the ,peal of1 #gh1 #bjd4 ,bergholt1 ,e41 #aaf1 #aai1 #abe4 ,betsy ,ross ,puzzle1 ,the1 #dj1 #agf4 ,bicycle ,thief1 ,the1 #f1 #aeb4 ,bishops--,guarded1 #hh1 #bad4 ---- in ,convocation1 #hi1 #bae4 ---- ,puzzle1 ,a ,new1 #ih1 #bbe4 ---- ,unguarded1 #hh1 #bad4 ,board1 ,the ,chess-1 #he4 ---- in ,compartments1 ,the1 #ajb1 #bbh4 ---- ,setting the1 #aje1 #bca4 ,boards with ,odd ,number of ,squares1 #hf1 #bab4 ,boat1 ,three ,men in a1 #gh1 #bjd4 ,bookworm1 ,the ,industrious1 #adc1 #bdh4 ,boothby1 ,guy1 #aed4 ,box1 ,the ,cardboard1 #di1 #aha4 ---- ,the ,paper1 #dj4 ,boys and ,girls1 #fg1 #aig4 ,bridges1 ,the ,monk and the1 #ge1 #bjb4 ,brigands1 ,the ,five1 #be1 #afd4 ,brocade1 ,the ,squares of1 #dg1 #ahj4 ,bun ,puzzle1 ,the1 #ce1 #agj4 ,busschop1 ,paul1 #agb4 ,buttons and ,string ,method1 #bcj4 ,cab ,numbers1 ,the1 #ae1 #aeg4 ,calendar ,puzzle1 ,a1 #adb1 #bdg4 ,canterbury ,puzzles1 ,the1 #ad1 #bh1 #eh1 #aag1 #aba1 #aie1 #bjb1 #bje1 #bjf1 #bab1 #bac1 #bag1 #bcc4 ,card ,frame ,puzzle1 ,the1 #aad1 #bch4 ---- ,magic ,squares1 #abc1 #bdd4 ---- ,players1 ,a ,puzzle for1 #gh1 #bjc4 ---- ,puzzle1 ,the ,t1 #aae1 #bci4 ---- ,triangles1 #aae1 #bci4 ,cards1 ,the ,cross of1 #aae1 #bch4 ,cardan1 #adb4 ,carroll1 ,lewis1 #dc4 ,castle ,treasure1 ,stealing the1 #aac1 #bcg4 ,cats1 the ,wizard's1 #db1 #agh4 ,cattle1 ,judkins's1 #f1 #aea4 ---- ,market1 ,at a1 #a1 #adh4 ,census ,puzzle1 ,a1 #g1 #aeb4 ,century ,puzzle1 ,the1 #af1 #aeh4 ---- ,the ,digital1 #af1 #aei4 ,chain ,puzzle1 ,a1 #add1 #bdi4 ---- ,the ,antiquary's1 #hc1 #bji4 ---- ,the ,cardboard1 #dj1 #agf4 ,change1 ,giving1 #d1 #aej4 ---- ,ways of giving1 #aea4 ,changing ,places1 #aj1 #aed4 ,channel ,island1 #ach4 ,charitable ,bequest1 ,a1 #b1 #adh ,charity1 ,indiscriminate1 #b1 #adh4 ,checkmate1 #ajg1 #bcc4 ,cheesemonger1 ,the ,eccentric1 #ff1 #aif4 ,chequered ,board ,divisions1 #he1 #baj4 ,cherries and ,plums1 #ef1 #ahi4 ,chess ,puzzles1 ,dynamical1 #if4 ---- ,statical1 #hh4 ---- ,various1 #aje4 ---- ,queer1 #ajg1 #bcc4 ,chessboard1 ,the1 #he4 ---- ,fallacy1 ,a1 #ada1 #bdg4 ---- ,guarded1 #ie4 ---- ,non-attacking ,arrangements1 #if4 ---- ,problems1 #hd4 ---- ,sentence1 ,the1 #hg1 #bad4 ---- ,solitaire1 #ajh1 #bcd4 ---- ,the ,chinese1 #hg1 #bac4 ---- ,the ,crowded1 #ia1 #bag4 ,chestnuts1 ,buying1 #f1 #aeb4 ,chinese ,money1 #d1 #aej4 ---- ,puzzle1 ,ancient1 #ajg1 #bcc4 ---- ---- ,the ,fashionable1 #dc4 ,christmas ,boxes1 ,the1 #d1 #aej4 ---- ,present1 ,mrs4 ,smiley's1 #df1 #agi4 ---- ,pudding1 ,the1 #dc1 #agh4 ,cigar ,puzzle1 ,the1 #aai1 #bdb4 ,circle1 ,the ,dissected1 #fi1 #aig4 ,cisterns1 ,how to ,make1 #ed1 #ahh4 ,civil ,service ,howler1 #aed4 ,clare1 ,john1 #eh4 ,clock ,formul1 #aed4 ---- ,puzzles1 #i4 ---- ,the ,club1 #aj1 #aed4 ---- ,the ,railway ,station1 #aa1 #aee4 ,clocks1 ,the ,three1 #aa1 #aed4 ,clothes ,line ,puzzle1 ,the1 #ej1 #ahb4 ,coast1 ,round the1 #fc1 #aie4 ,coincidence1 ,a ,queer1 #b1 #adh4 ,coins1 ,the ,broken1 #e1 #aej4 ---- ,the ,ten1 #eg1 #aij4 ---- ,two ,ancient1 #adj4 ,combination and ,group ,problems1 #gf4 ,compasses ,puzzle1 ,the1 #ec1 #ahf4 ,composite ,magic ,squares1 #abg1 #bdf4 ,cone ,puzzle1 ,the1 #ee1 #ahh4 ,corn1 ,reaping the1 #bj1 #afa4 ,cornfields1 ,farmer ,lawrence's1 #aja1 #bbg4 ,costermonger's ,puzzle1 ,the1 #f1 #aeb4 ,counter ,problems1 ,moving1 #eh4 ---- ,puzzle1 ,a ,new1 #ih1 #bbe4 ---- ,solitaire1 #ajg1 #bcd4 ,counters1 ,the ,coloured1 #ia1 #bag4 ---- ,the ,forty-nine1 #ib1 #bag4 ---- ,the ,nine1 #ad1 #aef4 ---- ,the ,ten1 #ae1 #aef4 ,crescent ,puzzle1 ,the1 #eb1 #ahd4 ,crescents of ,byzantium1 ,the ,five1 #ib1 #bai4 ,cricket ,match1 ,the ,village1 #aaf1 #bci4 ---- ,slow1 #aaf1 #bci4 ,cross and ,triangle1 #ce1 #afi4 ---- of ,cards1 #aae1 #bch4 ---- ,the ,folded1 #ce1 #afi4 ---- ,the ,southern1 #ic1 #bbj4 ,crosses1 ,counter1 #ha1 #bjg4 ---- from ,one1 ,two1 #ce1 #afh4 ---- ---- ,three1 #afi4 ,crossing ,river ,problems1 #aab4 ,crusader1 ,the1 #ajf1 #bcb4 ,cubes1 ,sums of1 #afe4 ,cushion ,covers1 ,the1 #df1 #agi4 ,cutting-out ,puzzle1 ,a1 #cg1 #agb4 ,cyclists' ,feast1 ,the1 #b1 #adh4 ,dairyman1 ,the ,honest1 #aaj1 #bce4 ,definition1 ,a ,question of1 #bc1 #afc4 ,de ,fonteney1 #aab4 ,deified ,puzzle1 ,the1 #gd1 #bjb4 ,delannoy1 #aab4 ,de ,morgan1 ,a41 #bg4 ,de ,tudor1 ,sir ,edwyn1 #ab1 #aee4 ,diabolique ,magic ,squares1 #abj4 ,diamond ,puzzle1 ,the1 #gd1 #bjb4 ,dice1 ,a ,trick with1 #aaf1 #bci4 ---- ,game1 ,the ,montenegrin1 #aai1 #bdb4 ---- ,numbers1 ,the1 #ag1 #afj4 ,die1 ,painting the1 #hd1 #baj1 ,digital ,analysis1 #aeg1 #aeh4 ---- ,division1 #af1 #aeh4 ---- ,multiplication1 #ae1 #aef4 ---- ,puzzles1 #ac4 ,digits1 ,adding the1 #af1 #aeh4 ---- and ,squares1 #ad1 #aee4 ---- ,odd and ,even1 #ad1 #aef4 ,dilemma1 ,an ,amazing1 #ajf1 #bcc4 ,diophantine ,problem1 #afd4 ,dissection ,puzzle1 ,an ,easy1 #ce1 #agj4 ---- ,puzzles1 #bg4 ---- ---- ,various1 #ce4 ,dividing ,magic ,squares1 #abd4 ,division1 ,digital1 #af1 #aeh4 ---- ,simple1 #bc1 #afc4 ,doctor's ,query1 ,the1 #aji1 #bce4 ,dogs ,puzzle1 ,the ,five1 #ib1 #bah4 ,domestic ,economy1 #e1 #aea4 ,domino ,frame ,puzzle1 ,the1 #aad1 #bch4 ,dominoes in ,progression1 #aad1 #bcg4 ---- ,the ,eighteen1 #abc1 #bde4 ---- ,the ,fifteen1 #hc1 #bji4 ---- ,the ,five1 #aad1 #bch4 ,donkey ,riding1 #ac1 #aee4 ,dormitory ,puzzle1 ,a1 #ha1 #bjh4 ,dovetailed ,block1 ,the1 #ade1 #bdi4 ,drayton's ,polyolbion1 #eh4 ,dungeon ,puzzle1 ,a1 #ig1 #bbd4 ,dungeons1 ,the ,siberian1 #abc1 #bdd4 ---- ,the ,spanish1 #abb1 #bdd4 ,dutchmen's ,wives1 ,the1 #bf1 #afg4 ,dynamical ,chess ,puzzles1 #if4 ,earth's ,girdle1 ,the1 #aci4 ,educational ,times ,reprints1 #bjd4 ,eggs1 ,a ,deal in1 #c1 #adi4 ---- ,obtaining the1 #adj4 ,election1 ,the ,muddletown1 #ai1 #afa4 ---- ,the ,parish ,council1 #ai1 #afa4 ,eleven1 ,the ,mystic1 #af1 #aei4 ,elopements1 ,the ,four1 #aac1 #bcg4 ,elrick1 ,e41 #bca4 ,engines1 ,the ,eight1 #fa1 #aid4 ,episcopal ,visitation1 ,an1 #ih1 #bbe4 ,estate1 ,farmer ,wurzel's1 #ea1 #ahd4 ,estates1 ,the ,yorkshire1 #ea1 #ahc4 ,euclid1 #ca1 #ach4 ,euler1 ,l41 #afe4 ,exchange ,puzzle1 ,the1 #ff1 #aif4 ,fallacy1 ,a ,chessboard1 #ada1 #bdg4 ,family ,party1 ,a1 #h1 #aec4 ,fare1 ,the ,passenger's1 #ac1 #aee4 ,farmer and his ,sheep1 ,the1 #bb1 #afc4 ,fence ,problem1 ,a1 #ba1 #afb4 ,fences1 ,the ,landowner's1 #db1 #agh4 ,fermat1 #afd1 #afh4 ,find the ,man's ,wife1 #adg1 #bea4 ,fly on the ,octahedron1 ,the1 #gj1 #aih4 ,fog1 ,mr4 ,gubbins in a1 #ah1 #afa4 ,football ,players1 ,the1 #aaf1 #bdj4 ,fraction1 ,a ,puzzling1 #ach4 ,fractions1 ,more ,mixed1 #af1 #aei4 ,frame ,puzzle1 ,the ,card1 #aad1 #bch4 ---- ---- ,the ,domino1 #aad1 #bch4 ,frankenstein1 ,e4,n41 #bcb4 ,frnicle1 ,b41 #aai1 #afh4 ,frogs1 ,the ,educated1 #ei1 #aid- ---- ,the ,four1 #ajc1 #bbi4 ---- ,the ,six1 #ei1 #aic4 ,frost1 ,a4,h41 #abj4 ,games1 ,puzzle1 #aag4 ---- ,problems concerning1 #aad4 ,garden1 ,lady ,belinda's1 #eb1 #ahf4 ---- ,puzzle1 ,the1 #di1 #ahb4 ,gardener and the ,cook1 ,the1 #adf1 #bea4 ,geometrical ,problems1 #bg4 ---- ,puzzles1 ,various1 #di4 ,george and the ,dragon1 ,st41 #aja1 #bbg4 ,getting ,upstairs1 ,such a1 #adc1 #bdh4 ,girdle1 the ,earth's1 #aci4 ,goat1 ,the ,tethered1 #ec1 #ahf4 ,grand ,lama's ,problem1 ,the1 #hf1 #bab4 ,grasshopper ,puzzle1 ,the1 #ei1 #aic4 ,greek ,cross ,puzzles1 #bh4 ---- ---- ,three from ,one1 #afi4 ,greyhound ,puzzle1 ,the1 #aja1 #bbg4 ,grocer and ,draper1 ,the1 #e1 #aea4 ,gros1 ,l41 #bdh4 ,group ,problems1 ,combination and1 #gf4 ,groups1 ,the ,three1 #ad1 #aef4 ,guarini1 #bbi4 ,hairdresser's ,puzzle1 ,the1 #acg4 ,halfpennies1 ,placing1 #adg1 #bea4 ,hampton ,court ,maze solved1 #acc4 ,hannah's ,puzzle1 #ge1 #bjb4 ,hastings1 ,the ,battle of1 #bc1 #afd4 ,hatfield ,maze solved1 #acf4 ,hat ,puzzle1 ,the1 #fg1 #aif4 ,hat-peg ,puzzle1 ,the1 #ic1 #bba4 ,hats1 ,the ,wrong1 #gh1 #bjc4 ,hay1 ,the ,trusses of1 #ah1 #afa4 ,heads or ,tails1 #bb1 #afc4 ,hearthrug1 ,mrs4 ,hobson's1 #cg1 #agb4 ,helmholtz1 ,von1 #da4 ,honey1 ,the ,barrels of1 #aaa1 #bcf4 ,honeycomb ,puzzle1 ,the1 #ge1 #bjb4 ,horse ,race ,puzzle1 ,the1 #aag1 #bdj4 ,horseshoes1 ,the ,two1 #dj1 #age4 ,houdin1 #fh4 ,hydroplane ,question1 ,the1 #ab1 #aee4 ,hymn-board ,poser1 ,the1 #ade1 #bej4 ,icosahedron ,puzzle1 ,the1 #gj1 #aih4 ,jack and the ,beanstalk1 #ade1 #bdi4 ,jackson1 ,john1 #ef4 ,jaenisch1 ,c4,f4 de1 #ib4 ,jampots1 ,arranging the1 #fh1 #aig4 ,jealous ,husbands1 ,five1 #aac1 #bcf4 ,joiner's ,problem1 ,the1 #cf1 #aga4 ---- ---- ,another1 #cg1 #aga4 ,jolly ,gaol-,birds1 ,eight1 #abb1 #bdc4 ---- ---- ,nine1 #abb1 #bdc4 ,journey1 ,the ,queen's1 #ajj1 #bbg4 ---- ,the ,rook's1 #if1 #bbd4 ,junior ,clerks' ,puzzle1 ,the1 #d1 #aej4 ,juvenile ,puzzle1 ,a1 #fh1 #aig4 ,kangaroos1 ,the ,four1 #ajb1 #bbh4 ,kelvin1 ,lord1 #da4 ,kennel ,puzzle1 ,the1 #aje1 #bca4 ,king and the ,castles1 ,the1 #ef1 #ahi4 ---- ,the ,forsaken1 #ajf1 #bcb4 ,kite-flying ,puzzle1 ,a1 #ed1 #ahg4 ,knight-guards1 ,the1 #ie1 #bbb4 ,knights1 ,king ,arthur's1 #gg1 #bjc4 ---- ,tour1 ,magic1 #abg1 #bdg4 ---- ---- ,the ,cubic1 #ajc1 #bbi4 ---- ---- ,the ,four1 #ajc1 #bbi4 ,labosne1 ,a41 #be1 #ij1 #baf4 ,labourer's ,puzzle1 ,the1 #ah1 #afj4 ,ladies' ,diary1 #bf4 ,lagrange1 ,j4,l41 #i4 ,laisant1 ,c4,a41 #gf4 ,lamp-posts1 ,painting the1 #ai1 #afa4 ,leap ,year1 #aee4 ---- ---- ,ladies1 ,the1 #ai1 #afa4 ,legacy1 ,a ,puzzling1 #bj1 #afa4 ,legal ,difficulty1 ,a1 #bc1 #afc4 ,le ,plongeon1 ,dr41 #bi4 ,letter ,block ,puzzle1 ,the1 #fj1 #aid4 ---- ,blocks1 ,the ,thirty-six1 #ia1 #baf4 ---- ,puzzle1 ,the ,fifteen1 #gi1 #bje4 ,level ,puzzle1 ,the1 #gd1 #bjb4 ,linoleum ,cutting1 #dh1 #aha4 ---- ,puzzle1 ,another1 #di1 #aha4 ,lion and the ,man1 ,the1 #ig1 #bbd4 ---- ,hunting1 #id1 #bbb4 ,lions and ,crowns1 #he1 #bab4 ---- ,the ,four1 #hh1 #bad4 ,lockers ,puzzle1 ,the1 #ad1 #aef4 ,locomotion and ,speed ,puzzles1 #aa4 ,lodging-house ,difficulty1 ,a1 #fa1 #aid4 ,london and ,wise1 #aca4 ,loyd1 ,sam1 #h1 #dc1 #dd1 #ih1 #add1 #bcb1 #bcc4 ,lucas1 ,edouard1 #af1 #gf1 #aab1 #aba4 ,luncheons1 ,the ,city1 #gg1 #bjc4 ,mac,mahon1 ,major1 #aji4 ,magic ,knight's ,tour1 #abg1 #bdg4 ---- ,square ,problems1 #aai4 ---- ---- ,card1 #abc1 #bdd4 ---- ---- of ,composites1 #abg1 #bdf4 ---- ---- of ,primes1 #abe4 ---- ---- of ,two ,degrees1 #abe1 #bde4 ---- ---- ,two ,new1 #abe1 #bde4 ---- ,strips1 #aba1 #bdc4 ,magics1 ,subtracting1 ,multiplying1 and ,dividing1 #abd4 ,maiden1 ,the ,languishing1 #ig1 #bbd4 ,mandarin's ,puzzle1 ,the1 #ajc1 #bcj4 ---- ,t ,puzzle1 ,the1 #abf1 #bdf4 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,weighing ,puzzles1 ,measuring1 ,packing1 and1 #aji4 ,wheels1 ,concerning1 #ee1 #ahh4 ,who was ,first8 #adb1 #bdg4 ,whyte1 ,w4,t41 #adg4 ,widow's ,legacy1 ,the1 #b1 #adh4 ,wife1 ,find the ,man's1 #adg1 #bea4 ,wilkinson1 ,rev4 ,mr41 #aic4 ,wilson1 ,professor1 #bi4 ,wilson's ,poser1 #i1 #aec4 ,wine and ,water1 #aaj1 #bce4 ---- ,the ,keg of1 #aaj1 #bce4 ,wotherspoon1 ,g41 #bdd4 ,yacht race1 ,the1 #ii1 #bbf4 ,youthful ,precocity1 #a1 #adh4 ,zeno1 #aci4 ,,the ,,end4