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[PDF page 19] CH. I] ARITHMETICAL RECREATIONS. 3 and 1790; doubtless these references are correct, but the following edi- tions, all of which I have seen, are the only ones of which I have any knowledge. In 1696 an edition was issued at Amsterdam. In 1723— six years after the death of Ozanam—one was issued in three volumes, with a supplementary fourth volume, containing (among other things) an appendix on puzzles: I believe that it would be difficult to find in any of the books current in England on mathematical amusements as many as a dozen puzzles which are not contained in one of these four volumes. Fresh editions were issued in 1741, 1750 (the second volume of which bears the date 1749), 1770, and 1790. The edition of 1750 is said to have been corrected by Montucla on condition that his name should not be associated with it; but the edition of 1790 is the earliest one in which reference is made to these corrections, though the editor is referred to only as Monsieur M***. Montucla expunged most of what was actually incorrect in the older editions, and added several historical notes, but unfortunately his scruples prevented him from striking out the accounts of numerous trivial experiments and truisms which over- load the work. An English translation of the original edition appeared in 1708, and I believe ran through four editions, the last of them being published in Dublin in 1790. Montucla’s revision of 1790 was translated by C. Hutton, and editions of this were issued in 1803, in 1814, and (in one volume) in 1840: my references are to the editions of 1803 and 1840. I proceed now to enumerate some of the elementary questions con- nected with numbers which for nearly three centuries have formed a large part of most compilations of mathematical amusements. They are given here mainly for their historical—not for their arithmetical— interest; and perhaps a mathematician may well omit them, and pass at once to the latter part of this chapter. These questions are of the nature of tricks or puzzles and I follow the usual course and present them in that form. I may note however that most of them are not worth proposing, even as tricks, unless either the modus operandi is disguised or the result arrived at is different from that expected; but, as I am not writing on conjuring, I refrain from alluding to the means of disguising the operations indicated, and give merely a bare enumeration of the steps essential to the success of the method used, though I may recall the fundamental rule that no trick, however good, will bear immediate repetition, and that, if it is

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[PDF page 20] 4 ARITHMETICAL RECREATIONS. [CH. I necessary to appear to repeat it, a different method of obtaining the result should be used. To find a number selected by some one. There are in- numerable ways of finding a number chosen by some one, provided the result of certain operations on it is known. I confine myself to methods typical of those commonly used. Any one acquainted with algebra will find no difficulty in modifying the rules here given or framing new ones of an analogous nature. First Method*. (i) Ask the person who has chosen the number to treble it. (ii) Enquire if the product is even or odd: if it is even, request him to take half of it; if it is odd, request him to add unity to it and then to take half of it. (iii) Tell him to multiply the result of the second step by 3. (iv) Ask how many integral times9 divides into the latter product: suppose the answer to ben. (v) Then the number thought of was 2n or 2n + 1, according as the result of step (i) was even or odd. The demonstration is obvious. Every even number is of the form 2n, and the successive operations applied to this give (i)6n, which is even; (ii) 1 26n = 3n; (iii) 3 × 3n = 9n; (iv) 1 99n = n; (v) 2n. Every odd number is of the form2n + 1, and the successive operations applied to this give (i) 6n + 3, which is odd; (ii) 1 2(6n + 3 + 1) = 3 n + 2; (iii) 3(3n + 2) = 9n + 6; (iv) 1 9(9n + 6) = n +a remainder; (v)2n + 1. These results lead to the rule given above. Second Method†. Ask the person who has chosen the number to perform in succession the following operations. (i) To multiply the number by 5. (ii) To add6 to the product. (iii) To multiply the sum by 4. (iv) To add9 to the product. (v) To multiply the sum by5. Ask to be told the result of the last operation: if from this product165 is subtracted, and then the remainder is divided by100, the quotient will be the number thought of originally. For let n be the number selected. Then the successive operations applied to it give (i) 5n; (ii) 5n + 6; (iii) 20n + 24; (iv) 20n + 33; (v) 100n + 165. Hence the rule. Third Method‡. Request the person who has thought of the num- ber to perform the following operations. (i) To multiply it by any number you like, say,a. (ii) To divide the product by any number, * Bachet, Problèmes plaisans, Lyons, 1624, problemi, p. 53. † A similar rule was given by Bachet, problemiv, p. 74. ‡ Bachet, problem v, p. 80.

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[PDF page 21] CH. I] ELEMENTARY TRICKS AND PROBLEMS. 5 say, b. (iii) To multiply the quotient byc. (iv) To divide this result by d. (v) To divide the final result by the number selected originally. (vi) To add to the result of operation (v) the number thought of at first. Ask for the sum so found: then, ifac/bd is subtracted from this sum, the remainder will be the number chosen originally. For, if n was the number selected, the result of the first four op- erations is to form nac/bd; operation (v) gives ac/bd; and (vi) gives n + (ac/bd), which number is mentioned. Butac/bd is known; hence, subtracting it from the number mentioned,n is found. Of coursea, b, c, d may have any numerical values it is liked to assign to them. For example, if a = 12, b = 4, c = 7, d = 3 it is sufficient to subtract7 from the final result in order to obtain the number originally selected. Fourth Method*. Ask some one to select a number less than90. (i) Request him to multiply it by10, and to add any number he pleases, a, which is less than10. (ii) Request him to divide the result of step (i) by 3, andtomentiontheremainder, say, b. (iii)Requesthimtomultiply the quotient obtained in step (ii) by10, and to add any number he pleases, c, which is less than10. (iv) Request him to divide the result of step (iii) by3, and to mention the remainder, sayd, and the third digit (from the right) of the quotient; suppose this digit ise. Then, if the numbers a, b, c, d, e are known, the original number can be at once determined. In fact, if the number is9x + y, where x ≯ 9 and y ≯ 8, and if r is the remainder whena − b + 3(c − d) is divided by 9, we have x = e, y = 9 − r. The demonstration is not difficult. For if the selected number is 9x+y, step(i)gives 90x+10y+a; (ii)let y+a = 3n+b, thenthequotient obtained in step (ii) is30x+3 y + n; step (iii) gives300x+30 y +10 n+ c; (iv) let n + c = 3 m + d, then the quotient obtained in step (iv) is 100x + 10y + 3n + m, which I will denote byQ. Now the third digit in Q must be x, because, since y ≯ 8 and a ≯ 9, we haven ≯ 5; and since n ≯ 5 and c ≯ 9, we havem ≯ 4; therefore 10y + 3n + m ≯ 99. Hence the third or hundreds digit inQ is x. Again, from the relations y + a = 3 n + b and n + c = 3 m + d, we have9m − y = a − b + 3(c − d): hence, if r is the remainder when a−b+3(c−d) is divided by9, we havey = 9−r. [This is always true, if we maker positive; but ifa−b+3(c−d) is negative, it is simpler to take y as equal to its numerical value; or we may prevent the occurrence of * Educational Times, London, May 1, 1895, vol.xl viii, p. 234.

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[PDF page 22] 6 ARITHMETICAL RECREATIONS. [CH. I this case by assigning proper values toa and c.] Thus x and y are both known, and therefore the number selected, namely9x + y, is known. Fifth Method *. Ask any one to select a number less than 60. (i) Request him to divide it by 3 and mention the remainder; sup- pose it to be a. (ii) Request him to divide it by4, and mention the remainder; suppose it to beb. (iii) Request him to divide it by5, and mention the remainder; suppose it to bec. Then the number selected is the remainder obtained by dividing40a + 45b + 36c by 60. This method can be generalized and then will apply to any number chosen. Let a′, b′, c′, . . . be a series of numbers prime to one another, and let p be their product. Let n be any number less thanp, and let a, b, c, . . .be the remainders when n is divided by a′, b′, c′, . . . respec- tively. Find a number A which is a multiple of the productb′c′d′ · · · and which exceeds by unity a multiple ofa′. Find a numberB which is a multiple ofa′c′d′ · · · and which exceeds by unity a multiple ofb′; and similarly find analogous numbersC, D, . . .. Rules for the calculation of A, B, C, . . .are given in the theory of numbers, but in general, if the numbers a′, b′, c′, . . .are small, the corresponding numbers,A, B, C, . . . can be found by inspection. I proceed to show thatn is equal to the remainder when Aa + Bb + Cc + · · · is divided by p. Let N = Aa+Bb+Cc +· · ·, and letM(x) stand for a multiple ofx. Now A = M(a′) + 1, therefore Aa = M(a′) + a. Hence, if the first term in N, that is Aa, is divided by a′, the remainder is a. Again, B is a multiple of a′c′d′ · · ·. Therefore Bb is exactly divisible by a′. Similarly Cc, Dd, . . . are each exactly divisible bya′. Thus every term in N, except the first, is exactly divisible bya′. Hence, if N is divided by a′, the remainder isa. But ifn is divided bya′, the remainder isa. Therefore N − n = M(a′) . Similarly N − n = M(b′) , N − n = M(c′) , ....................... But a′, b′, c′, . . .are prime to one another. ∴ N − n = M(a′b′c′ · · ·) = M(p) , that is, N = M(p) + n . * Bachet, problem vi, p. 84: Bachet added, on p. 87, a note on the previous history of the problem.

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[PDF page 23] CH. I] ELEMENTARY TRICKS AND PROBLEMS. 7 Now n is less thanp, hence ifN is divided byp, the remainder isn. The rule given by Bachet corresponds to the case ofa′ = 3, b′ = 4, c′ = 5, p = 60, A = 40, B = 45, C = 36. If the number chosen is less than 420, we may takea′ = 3, b′ = 4, c′ = 5, d′ = 7, p = 420, A = 280, B = 105, C = 336, D = 120. To find the result of a series of operations performed on any number (unknown to the questioner) without asking any questions. All rules for solving such problems ultimately depend on so arranging the operations that the number disappears from the final result. Four examples will suffice. First Example*. Request some one to think of a number. Suppose it to ben. Ask him (i) to multiply it by any number you please (say) a; (ii) then to add (say) b; (iii) then to divide the sum by (say) c. (iv) Next, tell him to takea/c of the number originally chosen; and (v) to subtract this from the result of the third operation. The result of the first three operations is(na + b)/c, and the result of operation (iv) isna/c: the difference between these isb/c, and therefore is known to you. For example, if a = 6, b = 12, c = 4, and a/c = 1 1 2, then the final result is 3. Second Example†. Ask A to take any number of counters that he pleases: suppose that he takesn counters. (i) Ask some one else, say B, to takep times as many, wherep is any number you like to choose. (ii) RequestA to giveq of his counters toB, whereq is any number you like to select. (iii) Next, askB to transfer toA a number of counters equal to p times as many counters asA has in his possession. Then there will remain inB’s handsq(p + 1)counters: this number is known to you; and the trick can be finished either by mentioning it or in any other way you like. The reason is as follows. The result of operation (ii) is thatB has pn + q counters, and A has n − q counters. The result of (iii) is that B transfers p(n − q) counters to A: hence he has left in his possession (pn + q) − p(n − q) counters, that is, he hasq(p + 1). For example, if originallyA took any number of counters, then (if you chose p equal to 2), first you would askB to take twice as many counters asA had done; next (if you choseq equal to3) you would ask * Bachet, problem viii, p. 102. † Bachet, problem xiii, p. 123: Bachet presented the above trick in a somewhat more general form, but one which is less effective in practice.

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[PDF page 24] 8 ARITHMETICAL RECREATIONS. [CH. I A to give 3 counters to B; and then you would askB to give to A a number of counters equal to twice the number then inA’s possession; after this was done you would know thatB had 3(2 + 1), that is,9 left. This trick (as also some of the following problems) may be per- formed equally well with one person, in which caseA may stand for his right hand and B for his left hand. Third Example. Ask some one to perform in succession the follow- ing operations. (i) Take any number of three digits. (ii) Form a new number by reversing the order of the digits. (iii) Find the difference of these two numbers. (iv) Form another number by reversing the order of thedigitsinthisdifference. (v)Addtogethertheresultsof(iii)and(iv). Then the sum obtained as the result of this last operation will be1089. An illustration and the explanation of the rule are given below. (i) 237 100 a + 10b + c (ii) 732 100 c + 10b + a (iii) 495 100( a − c − 1) + 90 + (10 + c − a) (iv) 594 100(10 + c − a) + 90 + ( a − c − 1) (v) 1089 900 + 180 + 9 Fourth Example*. The following trick depends on the same prin- ciple. Ask some one to perform in succession the following operations. (i) To write down any sum of money less than £12; the number of pounds not being the same as the number of pence. (ii) Toreverse this sum, that is, to write down a sum of money obtained from it by interchanging the numbers of pounds and pence. (iii) To find the differ- ence between the results of (i) and (ii). (iv) To reverse this difference. (v) To add together the results of (iii) and (iv). Then this sum will be £12. 18s. 11d. * Educational Times Reprints, 1890, vol.liii, p. 78.

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[PDF page 25] CH. I] ELEMENTARY TRICKS AND PROBLEMS. 9 For instance, take the sum £10. 17s. 5d.; we have £. s. d. (i) ........... 10 17 5 (ii) ........... 5 17 10 (iii) ........... 4 19 7 (iv) ........... 7 19 4 (v) ........... 12 18 11 Thefollowing work explainsthe rule, and shows thatthe finalresult is independent of the sum written down initially. £. s. d. (i) ........... a b c (ii) ........... c b a (iii) ........... a − c − 1 19 c − a + 12 (iv) ........... c − a + 12 19 a − c − 1 (v) ........... 11 38 11 The rule can be generalized to cover any system of monetary units. Problems involving Two Numbers. I proceed next to give a couple of examples of a class of problems which involve two numbers. First Example*. Suppose that there are two numbers, one even and the other odd, and that a person A is asked to select one of them, and that another person B takes the other. It is desired to know whetherA selected the even or the odd number. AskA to multi- ply his number by 2 (or any even number) andB to multiply his by3 (or any odd number). Request them to add the two products together and tell you the sum. If it is even, then originallyA selected the odd number, but if it is odd, then originallyA selected the even number. The reason is obvious. Second Example†. The above rule was extended by Bachet to any two numbers, provided they were prime to one another and one of them * Bachet, problem ix, p. 107. † Bachet, problem xi, p. 113.

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[PDF page 26] 10 ARITHMETICAL RECREATIONS. [CH. I was not itself a prime. Let the numbers bemand n, and suppose thatn is exactly divisible byp. Ask A to select one of these numbers, andB to take the other. Choose a number prime top, say q. Ask A to multiply his number by q, and B to multiply his number byp. Request them to add the products together and state the sum. Then A originally selected m or n, according as this result is not or is divisible byp. For example, m = 7, n = 15, p = 3, q = 2. Problems depending on the Scale of Notation. Many of the rules for finding two or more numbers depend on the fact that in arithmetic an integral number is denoted by a succession of digits, where each digit represents the product of that digit and a power of ten, and the number is equal to the sum of these products. For example, 2017 signifies (2 ×103)+(0 ×102)+(1 ×10)+7; that is, the2 represents 2 thousands, i.e. the product of2 and 103, the0 represents 0 hundreds, i.e.the product of0and 102; the1represents 1ten, i.e.the product of1 and 10, and the7 represents 7 units. Thus every digit has a local value. The application to tricks connected with numbers will be under- stood readily from three illustrative examples. First Example*. A common conjuring trick is to ask a boy among the audience to throw two dice, or to select at random from a box a domino on each half of which is a number. The boy is then told to recollect the two numbers thus obtained, to choose either of them, to multiply it by5, to add7 to the result, to double this result, and lastly to add to this the other number. From the number thus obtained, the conjurer subtracts 14, and obtains a number of two digits which are the two numbers chosen originally. For suppose that the boy selected the numbersa and b. Each of these is less than ten—dice or dominoes ensuring this. The successive operations give (i)5a; (ii)5a+7; (iii)10a+14; (iv)10a+14+ b. Hence, if 14 is subtracted from the final result, there will be left a number of two digits, and these digits are the numbers selected originally. An analogous trick might be performed in other scales of notation if it was thought necessary to disguise the process further. * Some similar questions were given by Bachet in problemxii, p. 117; by Oughtred in his Mathematicall Recreations (translated from or founded on van Etten’s work of 1633), London, 1653, problemxxxiv; and by Ozanam, parti, chapterx.

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[PDF page 27] CH. I] ELEMENTARY TRICKS AND PROBLEMS. 11 Second Example*. Similarly, if three numbers, say, a, b, c, are chosen, then, if each of them is less than ten, they can be found by the following rule. (i) Take one of the numbers, say,a, and multiply it by 2. (ii) Add3 to the product; the result is2a + 3. (iii) Multiply this by 5, and add 7 to the product; the result is10a + 22. (iv) To this sum add the second number. (v) Multiply the result by2. (vi) Add3 to the product. (vii) Multiply by5, and add the third number to the product. The result is100a + 10b + c + 235. Hence, if the final result is known, it is sufficient to subtract235 from it, and the remainder will be a number of three digits. These digits are the numbers chosen originally. I have seen a similar rule applied to determine the birthday and age of some one in the audience. The result is a number of six digits, of which the first two digits give the day of the month, the middle two digits the number of the month, and the last two digits the present age. Third Example†. The following rule for finding a man’s age is of the same kind. Take the tens digit of the year of birth; (i) multiply it by 5; (ii) to the product add2; (iii) multiply the result by2; (iv) to this product add the units digit of the birth-year; (v) subtract the sum from 110. The result is the man’s age in 1906. The algebraic proof of the rule is obvious. Leta and b be the tens and units digits of the birth-year. The successive operations give (i)5a; (ii) 5a + 2; (iii) 10a + 4 (iv) 10a + 4 + b; (v) 106 − (10a + b), which is his age in 1906. The rule can be easily adapted to give the age in any specified year. Other Problems with numbers in the denary scale.I may mention here two or three other slight problems dependent on the common scale of notation, which, as far as I am aware, are unknown to most compilers of books of puzzles. First Problem. The first of them is as follows. Take any number of three digits: reverse the order of the digits: subtract the number so formed from the original number: then, if the last digit of the difference is mentioned, all the digits in the difference are known. For leta be the hundreds digit of the number chosen,b be the tens digit, and c be the units digit. Therefore the number is100a + 10b + c. The number obtained by reversing the digits is100c + 10b + a. The * Bachet gave some similar questions in problemxii, p. 117. † A similar question was given by Laisant and Perrin in theirAlgèbre, Paris, 1892; and inL’Illustrationfor July 13, 1895.

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[PDF page 28] 12 ARITHMETICAL RECREATIONS. [CH. I difference of these numbers is equal to(100a+c)−(100c+a), that is, to 99(a − c). But a − c is not greater than9, and therefore the remainder can only be99, 198, 297, 396, 495, 594, 693, 792, or 891—in each case the middle digit being9 and the digit before it (if any) being equal to the difference between9 and the last digit. Hence, if the last digit is known, so is the whole of the remainder. Second Problem. The second problem is somewhat similar and is as follows. (i) Take any number; (ii) reverse the digits; (iii) find the difference between the number formed in (ii) and the given number; (iv) multiply this difference by any number you like to name; (v) cross out any digit except a nought; (vi) read the remainder. Then the sum of the digits in the remainder subtracted from the next highest multiple of nine will give the figure struck out. This follows at once from the fact that the result of operation (iii)— and therefore also of operation (iv)—is necessarily a multiple of nine, and it is known that the sum of the digits of every multiple of nine is itself a multiple of nine. Miscellaneous Questions. Besides these problems, properly so called, there are numerous questions on numbers which can be solved empirically, but which are of no special mathematical interest. As an instance I may quote a question which attracted some at- tention in London in 1893, and may be enunciated as follows. With the seven digits 9, 8, 7, 6, 5, 4, 0 express three numbers whose sum is 82: each digit, being used only once, and the use of the usual no- tations for fractions being allowed. One solution is 80.6˙9 + .7˙4 + .˙5. Similar questions are with the ten digits,9, 8, 7, 6, 5, 4, 3, 2, 1, 0, to express numbers whose sum is unity; a solution is35/70 and 148/296. If the sum were100, a solution would be50, 49, 1/2, and 38/76. A less straightforward question would be, with the nine digits,9, 8, 7, 6, 5, 4, 3, 2, 1, to express four numbers whose sum is100; a solution is 78, 15, 2√ 9, and 3√ 64. Problems with a series of things which are numbered. Any collection of things which can be distinguished one from the other—especially if numbered consecutively—afford easy concrete il- lustrations of questions depending on these elementary properties of numbers. As examples I proceed to enumerate a few familiar tricks. The first two of these are commonly shown by the use of awatch, the last three are best exemplified by the use of a pack of playing

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[PDF page 29] CH. I] ELEMENTARY TRICKS AND PROBLEMS. 13 cards, which readily lend themselves to such illustrations, and I present them in these forms. First Example*. The first of these examples is connected with the hours marked on the face of a watch. In this puzzle some one is asked to think of some hour, say,m, and then to touch a number that marks another hour, say,n. Then if, beginning with the number touched, he taps each successive hour marked on the face of the watch, going in the opposite direction to that in which the hands of the watch move, and reckoning to himself the taps asm, (m + 1), &c., the (n + 12)th tap will be on the hour he thought of. For example, if he thinks ofv and touchesix, then, if he taps successivelyix, viii, vii, vi, ..., going backwards and reckoning them respectively as5, 6, 7, 8, ..., the tap which he reckons as21 will be on thev. The reason of the rule is obvious, for he arrives finally at the (n + 12 − m)th hour from which he started. Now, since he goes in the opposite direction to that in which the hands of the watch move, he has to go over(n − m) hours to reach the hourm: also it will make no difference if in addition he goes over12 hours, since the only effect of this is to take him once completely round the circle. Now(n + 12− m) is always positive, since m < 12, and therefore if we make him pass over (n + 12− m) hours we can give the rule in a form which is equally valid whether m is greater or less thann. Second Example. The following is another well-known way of in- dicating on a watch-dial an hour selected by some one. I do not know who first invented it. If the hour is tapped by a pencil beginning at vii and proceeding backwards round the dial tovi, v, &c., and if the person who selected the number counts the taps, reckoning from the hour selected (thus, if he selectedx, he would reckon the first tap as the 11th), then the20th tap as reckoned by him will be on the hour chosen. For suppose he selected thenth hour. Then the8th tap is onxii and is reckoned by him as the(n + 8)th. The tap which he reckons as (n + 9)th is onxi, and generally the tap which he reckons as(n + p)th is on the hour(20 − p). Hence, putting p − 20 − n, the tap which he reckons as20th is on the hourn. Of course the hours indicated by the first seven taps are immaterial. * Bachet, problem xx, p. 155; Oughtred, Mathematicall Recreations, London, 1653, p. 28.
