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[PDF page 30] 14 ARITHMETICAL RECREATIONS. [CH. I Extension. It is obvious that the same trick can be performed with any collection ofm things, such as cards or dominoes, which are distinguishable one from the other, provided m < 20. For suppose the m things are arranged on a table in some numerical order, and the nth thing is selected by a spectator. Then the first(19 − m) taps are immaterial, the(20 − m)th tap must be on themth thing and be reckoned by the spectator as the(n+20 − m)th, the(20 − m+1)th tap must be on the(m−1)th thing and be reckoned as the(n+20−m+1)th, and finally the(20 − n)th tap will be on thenth thing and is reckoned as the 20th tap. Third Example. The following example rests on an extension of the method used in the last question; it is very simple, but I have never seen it previously described in print. Suppose that a pack ofn cards is given to some one who is asked to select one out of the firstm cards and to remember (but not to mention) what is its number from the top of the pack (say it is actually thexth card in the pack). Then take the pack, reverse the order of the topm cards (which can be easily effected by shuffling), and transfery cards (where y < n − m) from the bottom to the top of the pack. The effect of this is that the card originally chosen is now the(y + m − x + 1)th from the top. Return to the spectator the pack so rearranged, and ask that the top card be counted as the(x + 1)th, the next as the(x + 2)th, and so on, in which case the card originally chosen will be the(y + m + 1)th. Now y and m can be chosen as we please, and may be varied every time the trick is performed; thus any one unskilled in arithmetic will not readily detect the modus operandi. Fourth Example*. Place a card on the table, and on it place as many other cards from the pack as with the number of pips on the card will make a total of twelve. For example, if the card placed first on the table is the five of clubs, then seven additional cards must be placed on it. The court cards may have any values assigned to them, but usually they are reckoned as tens. This is done again with another card, and thus another pile is formed. The operation may be repeated either only three or four times or as often as the pack will permit of such piles being formed. If finally there are p such piles, and if the number of cards left over isr, then the sum of the number of pips on the bottom cards of all the piles will be13(p − 4) + r. * A particular case of this problem was given by Bachet, problemxvii, p. 138.

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[PDF page 31] CH. I] MEDIEV AL PROBLEMS IN ARITHMETIC. 15 For, if x is the number of pips on the bottom card of a pile, the number of cards in that pile will be13 − x. A similar argument holds for each pile. Also there are 52 cards in the pack; and this must be equal to the sum of the cards in thep piles and ther cards left over. ∴ (13 − x1) + (13 − x2) + · · · + (13 − xp) + r = 52 , ∴ 13p − (x1 + x2 + · · · + xp) + r = 52 , ∴ x1 + x2 + · · · + xp = 13p − 52 + r = 13(p − 4) + r . More generally, if a pack ofn cards is taken, and if in each pile the sum of the pips on the bottom card and the number of cards put on it is equal tom, then the sum of the pips on the bottom cards of the piles will be(m + 1)p + r − n. In an écarté pack n = 32, and it is convenient to take m = 15. Fifth Example. It may be noticed that cutting a pack of cards never alters the relative position of the cards provided that, if necessary, we regard the top card as following immediately after the bottom card in the pack. This is used in the following trick*. Take a pack, and deal the cards face upwards on the table, calling them one, two, three, &c. as you put them down, and noting in your own mind the card first dealt. Ask some one to select a card and recollect its number. Turn the pack over, and let it be cut (not shuffled) as often as you like. Enquire what was the number of the card chosen. Then, if you deal, and as soon as you come to the original first card begin (silently) to count, reckoning this as one, the selected card will appear at the number mentioned. Of course, if all the cards are dealt before reaching this number, you must turn the cards over and go on counting continuously. Another similar trick is performed by handing the pack face up- wards to some one, and asking him to select a card and state its num- ber, reckoning from the top; suppose it to be thenth. Next, ask him to choose a number at which it shall appear in the pack; suppose he selects themth. Take the pack and secretly movem − n cards from the bottom to the top (or ifn is greater thanm, then n − m from the top to the bottom) and of course the card will be in the required position. * Bachet, problem xix, p. 152.

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[PDF page 32] 16 ARITHMETICAL RECREATIONS. [CH. I Medieval Problems in Arithmetic. Before leaving the sub- ject of these elementary questions, I may mention a few problems which for centuries have appeared in nearly every collection of mathemati- cal recreations, and therefore may claim what is almost a prescriptive right to a place here. First Example*. The following is a sample of one class of these puzzles. Three men robbed a gentleman of a vase, containing24 ounces of balsam. Whilst running away they met in a wood with a glass-seller, of whom in a great hurry they purchased three vessels. On reaching a place of safety they wished to divide the booty, but they found that their vessels contained 5, 11, and 13 ounces respectively. How could they divide the balsam into equal portions? Problems like this can be worked out only by trial: there are several solutions, of which one is as follows. The vessels can contain...... 24 oz. 13 oz. 11 oz. 5 oz. Their contents originally are 24 ... 0 ... 0 ... 0 ... First, make their contents ... 0 ... 8 ... 11 ... 5 ... Second, ” ” ... 16 ... 8 ... 0 ... 0 ... Third, ” ” ... 16 ... 0 ... 8 ... 0 ... Fourth, ” ” ... 3 ... 13 ... 8 ... 0 ... Fifth, ” ” ... 3 ... 8 ... 8 ... 5 ... Sixth, ” ” ... 8 ... 8 ... 8 ... 0 ... Second Example†. The next of these is a not uncommon game, played by two people, say A and B. A begins by mentioning some number not greater than (say) six,B may add to that any number not greater than six,A may add to that again any number not greater than six, andsoon. Hewinswhoisthefirsttoreach(say) 50. Obviously, ifA calls 43, then whateverB adds to that,A can win next time. Similarly, if A calls 36, B cannot prevent A’s calling43 the next time. In this way it is clear that the key numbers are those forming the arithmetical progression 43, 36, 29, 22, 15, 8, 1; and whoever plays first ought to win. Similarly, if no number greater thanm may be added at any one time, andn is the number to be called by the victor, then the key num- * Some similar problems were given by Bachet, appendix, problem iii, p. 206; problem ix, p. 233; by Oughtred in hisRecreations, p. 22: and by Ozanam, 1803 edition, vol. i, p. 174; 1840 edition, p. 79. Earlier instances occur in Tartaglia’s writings. † Bachet, problem xxii, p. 170.

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[PDF page 33] CH. I] MEDIEV AL PROBLEMS IN ARITHMETIC. 17 bers will be those forming the arithmetical progression whose common difference is m + 1 and whose smallest term is the remainder obtained by dividing n by m + 1. The same game may be played in another form by placingn coins, matches, or other objects on a table, and directing each player in turn to take away not more thanm of them. Whoever takes away the last coin wins. Obviously the key numbers are multiples ofm + 1, and the first player who is able to leave an exact multiple of(m + 1) coins can win. Perhaps a better form of the game is to make that player lose who takes away the last coin, in which case each of the key numbers exceeds by unity a multiple ofm + 1. Mr Loyd has also suggested* a modification which is equivalent to placing n counters in the form of a circle, and allowing each player in succession to take away not more thanmof them which are in unbroken sequence: m being less thann and greater than unity. In this case the second of the two players can always win. Recent Extension of this Problem. The games last described are very simple, but if we impose on the original problem the additional restriction that each player may not add the same number more than three times, the analysis becomes by no means easy. It is difficult in this case to say whether it is an advantage to begin or not. I have never seen this extension described in print, and I will therefore enunciate it at length. Suppose that each player is given eighteen cards, three of them marked 6, three marked 5, three marked 4, three marked 3, three marked 2, and three marked 1. They play alternately; A begins by playing one of his cards; thenB plays one of his, and so on. He wins who first plays a card which makes the sum of the points or numbers on all the cards played exactly equal to50, but he loses if he plays a card which makes this sum exceed50. The game can be played men- tally or by noting the numbers on a piece of paper, and in practice it is unnecessary to use cards. Thus, if they play as followsA, 4; B, 3; A, 1; B, 6; A, 3; B, 4; A, 4; B, 5; A, 4; B, 4; A, 5; the game stands at43. B can now win, for he may safely play3, since A has not another4 wherewith to follow it; and ifA plays less than4, B will win the next time. Again, if they play * Tit-Bits, London, July 17, Aug. 7, 1897.

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[PDF page 34] 18 ARITHMETICAL RECREATIONS. [CH. I thus, A, 6; B, 3; A, 1; B, 6; A, 3; B, 4; A, 2; B, 5; A, 1; B, 5; A, 2; B, 5; A, 2; B, 3; A is now forced to play1, and B wins by playing1. The game can be also played if each player is given only two cards of each kind. Third Example. The following medieval problem is somewhat more elaborate. Suppose that three people, P, Q, R, select three things, which we may denote by a, e, i, respectively, and that it is desired to find by whom each object was selected*. Place 24 counters on a table. AskP to take one counter,Q to take two counters, andR to take three counters. Next, ask the person who selected a to take as many counters as he has already, whoever selected etotaketwiceasmanycountersashehasalready, andwhoeverselected i to take four times as many counters as he has already. Note how many counters remain on the table. There are only six ways of distributing the three things among P, Q, and R; and the number of counters remaining on the table is different for each way. The remainders may be 1, 2, 3, 5, 6, or 7. Bachet summed up the results in the mnemonic linePar fer (1) César (2) jadis (3) devint (5) si grand(6) prince (7). Corresponding to any remainder is a word or words containing two syllables: for instance, to the remainder 5 corresponds the word devint. The vowel in the first syllable indicates the thing selected byP, the vowel in the second syllable indicates the thing selected byQ, and of courseR selected the remaining thing. Salve certa animae semita vita quieswas suggested by Oughtred† as an alternative mnemonic line. Extension. M. Bourlet, in the course of a very kindly notice‡ of the second edition of this work, has given a much neater solution of the above question, and has extended the problem to the case ofn people, P0, P1, P2, . . . , Pn−1, each of whom selects one object, out of a collection of n objects, such as dominoes or cards. It is required to know which domino or card was selected by each person. Let us suppose the dominoes to be denoted or marked by the num- bers 0, 1, . . . , n− 1, instead of by vowels. Give one counter toP1, two counters to P2, and generally k counters to Pk. Note the number of counters left on the table. Next ask the person who had chosen the * Bachet, problem xxv, p. 187. † Mathematicall Recreations, London, 1653, p. 20. ‡ Bulletin des sciences mathématiques, Paris, 1893, vol.xvii, pp. 105–107.

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[PDF page 35] CH. I] MEDIEV AL PROBLEMS IN ARITHMETIC. 19 domino 0 to take as many counters as he had already, and generally whoever had chosen the dominoh to take nh times as many dominoes as he had already: thus ifPk had chosen the domino numberedh, he would take nhk counters. Note the total number of counters taken, i.e. P nhk. Divide it by n, then the remainder will be the number on the domino selected byP0; divide the quotient byn, and the remainder will be the number on the domino selected byP1; divide this quotient by n, and the remainder will be the number on the domino selected by P2; and so on. In other words, if the number of counters taken is expressed in the scale of notation whose radix isn, then the(h + 1)th digit from the right will give the number on the domino selected byPh. Thus in Bachet’s problem with3 people and3 dominoes, we should first give one counter toQ, and two counters toR, while P would have no counters; then we should ask the person who selected the domino marked 0 or a to take as many counters as he had already, whoever selected the domino marked1 or e to take three times as many counters as he had already, and whoever selected the domino marked2 or i to take nine times as many counters as he had already. By noticing the original number of counters, and observing that3 of these had been given to Q and R, we should know the total number taken byP, Q, and R. If this number were divided by3, the remainder would be the number of the domino chosen byP; if the quotient were divided by3 the remainder would be the number of the domino chosen byQ; and the final quotient would be the number of the domino chosen byR. I may add that Bachet also discussed the case whenn = 4, which had been previously considered by Diego Palomino in 1599, but as M. Bourlet’s method is general, it is unnecessary to discuss further particular cases. Decimation. ThelastoftheseantiqueproblemstowhichIreferred consists in placing men round a circle so that if everynth man is killed the remainder shall be certain specified individuals. When decimation was a not uncommon punishment a knowledge of this kind may have had practical interest. Hegesippus* says that Josephus saved his life by such a device. Ac- cording to his account, after the Romans had captured Jotopat, Jose- phus and forty other Jews took refuge in a cave. Josephus, much to his disgust, found that all except himself and one other man were resolved * De Bello Judaico, bk. iii, chaps. 16–18.

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[PDF page 36] 20 ARITHMETICAL RECREATIONS. [CH. I to kill themselves, so as not to fall into the hands of their conquerors. Fearing to show his opposition too openly he consented, but declared that the operation must be carried out in an orderly way, and sug- gested that they should arrange themselves round a circle and that every third person should be killed until but one man was left, who must then commit suicide. It is alleged that he placed himself and the other man in the31st and 16th place respectively, with a result which will be easily foreseen. The question is usually presented in the following form. A ship, carrying as passengers fifteen Turks and fifteen Christians, encountered a storm, and the pilot declared that, in order to save the ship and crew, one-half of the passengers must be thrown into the sea. To choose the victims the passengers were placed round a circle, and it was agreed thateveryninthmanshouldbecastoverboard, reckoningfromacertain point. It is desired to find an arrangement by which all the Christians should be saved.* Problems like this can be easily solved by counting, but it is im- possible to give a general rule. In this case, the Christians, reckoning from the man first counted, must occupy the places1, 2, 3, 4, 10, 11, 13, 14, 15, 17, 20, 21, 25, 28, 29. This arrangement can be recollected by the positions of the vowels in the following doggerel rhyme, From numbers’ aid and art, never will fame depart, where a stands for 1, e for 2, i for 3, o for 4, and u for 5. Hence (looking only at the vowels in the verse) the order is4 Christians, 5 Turks, 2 Christians, 1 Turk, 3 Christians, 1 Turk, 1 Christian, 2 Turks, 2 Christians, 3 Turks, 1 Christian, 2 Turks, 2 Christians, 1 Turk. Other similar mnemonic lines in French and in Latin were given by Bachet and by Ozanam respectively. Arithmetical Fallacies. I insert next some instances of demonstrations† leading to arithmetical results which are obviously * Bachet, problem xxiii, p. 174. The same problem had been previously enunci- ated by Tartaglia. † Of the fallacies given in the text, the first, second, and third, are well known; the fourth is not new, but the earliest work in which I recollect seeing it is my Algebra, Cambridge, 1890, p. 430; the fifth is given in G.C. Chrystal’sAlgebra,

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[PDF page 37] CH. I] ARITHMETICAL F ALLACIES. 21 impossible. I include algebraical proofs as well as arithmetical ones. The fallacies are so patent that in preparing the first and second editions I did not think such questions worth printing, but, as some correspondents have expressed a contrary opinion, I give them for what they are worth. First fallacy. One of the oldest of these—and not a very interest- ing specimen—is as follows. Suppose thata = b, then ab = a2 . ∴ ab − b2 = a2 − b2 . ∴ b(a − b) = (a + b)(a − b) . ∴ b = a + b . ∴ b = 2b . ∴ 1 = 2 . Second Fallacy. Another instance, almost as puerile, is as follows. Let a and b be two unequal numbers, and let c be their arithmetic mean, hence a + b = 2c . ∴ (a + b)(a − b) = 2c(a − b) . ∴ a2 − 2ac = b2 − 2bc . ∴ a2 − 2ac + c2 = b2 − 2bc + c2 . ∴ (a − c)2 = (b − c)2 . ∴ a = b . Edinburgh, 1889, vol. ii, p. 159; the eighth is due to G.T. Walker, and, as far as I know, has not appeared in any other book; the ninth is due to D’Alembert; and the tenth to F. Galton. A mechanical demonstration that1 = 2 was given by R. Chartres inKnowledge, July, 1891. J.L.F. Bertrand pointed out that a demonstration that 1 = −1 can be also obtained from the proposition in the Integral Calculus that, if the limits are constant, the order of integration is indifferent; hence the integral to x (from x = 0 to x = 1) of the integral to y (from y = 0 to y = 1) of a function φ should be equal to the integral toy (from y = 0 to y = 1) of the integral tox (from x = 0 to x = 1) of φ, but if φ = (x2 − y2)/(x2 + y2)2, this gives 1 4 π = − 1 4 π.

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[PDF page 38] 22 ARITHMETICAL RECREATIONS. [CH. I Third Fallacy. Another example, the idea of which is due to John Bernoulli, may be stated as follows. We have (−1)2 = 1 . Take logarithms, ∴ 2 log(−1) = log 1 = 0 . ∴ log(−1) = 0 . ∴ −1 = e0 . ∴ −1 = 1 . The same argument may be expressed thus. Letx be a quantity which satisfies the equation ex = −1 . Square both sides, ∴ e2x = 1 . ∴ 2x = 0 . ∴ x = 0 . ∴ ex = e0 . But ex = −1 and e0 = 1, ∴ −1 = 1 . Fourth Fallacy. As yet another instance, we know that log(1 + x) = x − 1 2 x2 + 1 3 x3 − · · · . If x = 1, the resulting series is convergent; hence we have log 2 = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 + 1 9 − · · · . ∴ 2 log 2 = 2− 1 + 2 3 − 1 2 + 2 5 − 1 3 + 2 7 − 1 4 + 2 9 − · · · . Taking those terms together which have a common denominator, we obtain 2 log 2 = 1 +1 3 − 1 2 + 1 5 + 1 7 − 1 4 + 1 9 − · · · = 1 − 1 2 + 1 3 − 1 4 + 1 5 − · · · = log 2. Hence 2 = 1 .

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[PDF page 39] CH. I] ARITHMETICAL F ALLACIES. 23 Fifth Fallacy. This fallacy is very similar to that last given. We have log 2 = 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + · · · =   1 + 1 3 + 1 5 + · · ·  −   1 2 + 1 4 + 1 6 + · · ·  =   1 + 1 3 + 1 5 + · · ·  +   1 2 + 1 4 + 1 6 + · · ·  − 2   1 2 + 1 4 + 1 6 + · · ·  =  1 + 1 2 + 1 3 + · · ·

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−   1 + 1 2 + 1 3 + · · ·  = 0 . The error in each of the foregoing examples is obvious, but the fallacies in the next examples are concealed somewhat better. Sixth Fallacy. We can write the identity√−1 = √−1 in the form r −1 1 = r 1 −1 , hence √−1√ 1 = √ 1√−1 , therefore ( √ −1)2 = ( √ 1)2 , that is, −1 = 1 . Seventh Fallacy. Again, we have √a × √ b = √ ab . Hence √ −1 × √ −1 = p (−1)(−1) , therefore ( √ −1)2 = √ 1 , that is, −1 = 1 . Eighth Fallacy. The following demonstration depends on the fact that an algebraical identity is true whatever be the symbols used in it, and it will appeal only to those who are familiar with this fact. We have, as an identity, √x − y = i√y − x ....... (i), where i stands either for +√−1 or for −√−1. Now an identity in x and y is necessarily true whatever numbers x and y may represent. First put x = a and y = b, ∴ √ a − b = i √ b − a .......(ii).

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[PDF page 40] 24 ARITHMETICAL RECREATIONS. [CH. I Next put x = b and y = a, ∴ √ b − a = i √ a − b ...... (iii). Also since (i) is an identity, it follows that in (ii) and (iii) the symboli must be the same, that is, it represents+√−1 or −√−1 in both cases. Hence, from (ii) and (iii), we have √ a − b √ b − a = i2√ b − a √ a − b , ∴ 1 = i2 , that is 1 = −1 . Ninth Fallacy. The following fallacy is due to D’Alembert*. We know that if the product of two numbers is equal to the product of two other numbers, the numbers will be in proportion, and from the definition of a proportion it follows that if the first term is greater than the second, then the third term will be greater than the fourth: thus, if ad = bc, then a : b = c : d, and if in this proportiona > b , then c > d. Now if we puta = d = 1 and b = c = −1 we have four numbers which satisfy the relationad = bc and such thata > b ; hence, by the proposition, c > d , that is, −1 > 1, which is absurd. Tenth Fallacy. The mathematical theory of probability leads to various paradoxes: of these one specimen† will suffice. Suppose three coins to be thrown up and the fact whether each comes down head or tail to be noticed. The probability that all three coins come down head is clearly( 1 2)3, that is, is1 8; similarly the probability that all three come down tail is1 8: hence the probability that all the coins come down alike (i.e.either all of them heads or all of them tails) is1 4. But, of three coins thus thrown up, at least two must come down alike; now the probability that the third coin comes down head is1 2 and the probability that it comes down tail is1 2, thus the probability that it comes down the same as the other two coins is 1 2: hence the probability that all the coins come down alike is 1 2. I leave to my readers to say whether either of these conflicting conclusions is right and if so, which. Arithmetical Problems. To the above examples I may add the fol- lowing questions, which I have often propounded in past years: though not fallacies, they may serve to illustrate the fact that the answer to * Opuscules mathématiques, Paris, 1761, vol.i, p. 201. † See Nature, Feb. 15, March 1, 1894, vol.xlix, pp. 365–366, 413.

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[PDF page 41] CH. I] ARITHMETICAL F ALLACIES. 25 an arithmetical question is frequently different to what a hasty reader might suppose. The first of these questions is as follows. Two clerks are engaged, one at a salary commencing at the rate of (say) £100 a year with a rise of £20 every year, the other at a salary commencing at the same rate (£100 a year) with a rise of £5 every half-year, in each case payments being made half-yearly: which has the larger income? The answer is the latter; for in the first year the first clerk receives £100, but the second clerk receives £50 and £55 as his two half-yearly payments and thus receives in all £105. In the second year the first clerk receives £120, but the second clerk receives £60 and £65 as his two half-yearly payments and thus receives in all £125. In fact the second clerk will always receive £5 a year more than the first clerk. As another question take the following. A man bets1/nth of his money on an even chance (say tossing heads or tails with a penny): he repeats this again and again, each time betting1/nth of all the money then in his possession. If, finally, the number of times he has won is equal to the number of times he has lost, has he gained or lost by the transaction? He has, in fact, lost. Here is another simple question to which not unfrequently I have received incorrect answers. One tumbler is half-full of wine, another is half-full of water: from the first tumbler a teaspoonful of wine is taken out and poured into the tumbler containing the water: a teaspoonful of the mixture in the second tumbler is then transferred to the first tumbler. As the result of this double transaction, is the quantity of wine removed from the first tumbler greater or less than the quantity of water removed from the second tumbler? Nineteen people out of twenty will say it is greater, but this is not the case. Routes on a Chess-Board. A not uncommon problem can be gen- eralised as follows*. Construct a rectangular board ofmn cells (or small squares) by rulingm + 1 vertical lines andn + 1 horizontal lines. It is required to know how many routes can be taken from the top left-hand corner to the bottom right-hand corner, the motion being along the ruled lines and its direction being always either vertically downwards or horizontally from left to right. The answer is(m+ n)!/m!n!: thus on a square board containing16 cells (i.e. one-quarter of a chess-board), * The substance of the problem was given in a scholarship paper set at Cambridge about 30 years ago, and possibly was not new then.
