Mathematical Recreations and Essays · W. W. Rouse Ball
The interest excited by statements of the relations between numbers of certain forms has been often remarked. The majority of works on mathematical recreations include several such problems, which are obvious to any one acquainted with the elements of algebra, but which to many who are ignorant of that subject possess the same kind of charm that some mathematicians find in the more recondite propositions of higher arithmetic. I shall devote the bulk of this chapter to these elementary problems, but I append a few remarks on one or two questions in the theory of numbers.
Before entering on the subject of the chapter, I may add that a large proportion of the elementary questions mentioned here and in the following two chapters are taken from one of two sources. The first of these is the classical Problèmes plaisans et délectables, by C.G. Bachet, sieur de Méziriac, of which the first edition was published in 1612 and the second in 1624: it is to the edition of 1624 that the references hereafter given apply. Several of Bachet's problems are taken from the writings of Alcuin, Pacioli di Burgo, Tartaglia, or Cardan, and possibly some of them are of oriental origin, but I have made no attempt to add such references. The other source to which I alluded above is Ozanam's Récréations mathématiques et physiques. The greater portion of the original edition, published in two volumes at Paris in 1694, was a compilation from the works of Bachet, Leurechon, Mydorge,
van Etten, and Oughtred: this part is excellent, but the same cannot be said of the additions due to Ozanam. In the Biographie Universelle allusion is made to subsequent editions issued in 1720, 1735, 1741, 1778, and 1790; doubtless these references are correct, but the following editions, 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 overload 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 connected 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 necessary to appear to repeat it, a different method of obtaining the result should be used.
[Elementary Questions on Numbers (Miscellaneous)] [Elementary Tricks and Problems]To find a number selected by some one There are
innumerable 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/ (Bachet, Problèmes plaisans, Lyons, 1624, problem i, p. 53.) (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 times 9 divides into the latter product: suppose the answer to be n. (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 form
2n + 1 , and the successive operations applied to this give (i) 6n + 3 , which is odd; (ii) 1/2(6n + 3 + 1) = 3n + 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/ (A similar rule was given by Bachet, problem iv, p. 74.) Ask the person who has chosen the number to perform in succession the following operations. (i) To multiply the number by 5. (ii) To add 6 to the product. (iii) To multiply the sum by 4. (iv) To add 9 to the product. (v) To multiply the sum by 5. Ask to be told the result of the last operation: if from this product 165 is subtracted, and then the remainder is divided by 100 , 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/ (Bachet, problem v, p. 80.) Request the person who has thought of the number to perform the following operations. (i) To multiply it by any number you like, say, a. (ii) To divide the product by any number, say, b. (iii) To multiply the quotient by c. (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, if ac/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 operations is to form nac/bd ; operation (v) gives ac/bd ; and (vi) gives n + (ac/bd) , which number is mentioned. But ac/bd is known; hence, subtracting it from the number mentioned, n is found. Of course a , 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 subtract 7 from the final result in order to obtain the number originally selected.
Fourth Method/ (Educational Times, London, May 1, 1895, vol. xlviii, p. 234.) Ask some one to select a number less
than 90. (i) Request him to multiply it by 10 , and to add any number he pleases, a , which is less than 10. (ii) Request him to divide the result of step (i) by 3 , and to mention the remainder, say, b. (iii) Request him to multiply the quotient obtained in step (ii) by 10 , and to add any number he pleases, c , which is less than 10. (iv) Request him to divide the result of step (iii) by 3 , and to mention the remainder, say d , and the third digit (from the right) of the quotient; suppose this digit is e. Then, if the numbers a , b , c , d , e are known, the original number can be at once determined. In fact, if the number is 9x + y , where x 9 and y 8 , and if r is the remainder when a-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 , then the quotient obtained in step (ii) is 30x + 3y + n ; step (iii) gives 300x + 30y + 10n + c ; (iv) let n + c = 3m + d , then the quotient obtained in step (iv) is 100x+ 10y+ 3n + m , which I will denote by Q. Now the third digit in Q must be x , because, since y 8 and a 9 , we have n 5 ; and since n 5 and c 9 , we have m 4 ; therefore 10y + 3n + m 99. Hence the third or hundreds digit in Q is x.
Again, from the relations y + a = 3n + b and n + c = 3m + d , we have 9m-y = a-b + 3(c-d) : hence, if r is the remainder when a-b +3(c-d) is divided by 9 , we have y=9-r. [This is always true, if we make r positive; but if a-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 this case by assigning proper values to a and c .] Thus x and y are both known, and therefore the number selected, namely 9x + y , is known.
Fifth Method/ (Bachet, problem vi, p. 84: Bachet added, on p. 87, a note on the previous history of the problem.) Ask any one to select a number less than 60. (i) Request him to divide it by 3 and mention the
remainder; suppose it to be a. (ii) Request him to divide it by 4 , and mention the remainder; suppose it to be b. (iii) Request him to divide it by 5 , and mention the remainder; suppose it to be c. Then the number selected is the remainder obtained by dividing 40a + 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 than p , and let a, b, c... be the remainders when n is divided by a', b', c'... respectively. Find a number A which is a multiple of the product b'c'd' and which exceeds by unity a multiple of a'. Find a number B which is a multiple of a'c'd' and which exceeds by unity a multiple of b' ; and similarly find analogous numbers C, 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 that n is equal to the remainder when Aa + Bb + Cc +... is divided by p.
Let N = Aa + Bb + Cc +... , and let M(x) stand for a multiple of x.
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 by a'. Thus every term in N , except the first, is exactly divisible by a'. Hence, if N is divided by a' , the remainder is a. But if n is divided by a' , the remainder is a. Therefore N-n = M(a'). ; Similarly N-n = M(b'), ; N-n = M(c'), ; 2 ; But $a', b', c'...$ are prime to one another.; Therefore N-n = M(a'b'c')$ = M(p),$ ; that is, N = M(p) + n.;
Now n is less than p , hence if N is divided by p , the remainder is n.
The rule given by Bachet corresponds to the case of a' = 3 , b'= 4 , c' = 5 , p = 60 , A = 40 , B = 45 , C = 36. If the number chosen is less than 420, we may take a' = 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/ (Bachet, problem viii, p. 102.) Request some one to think of a number. Suppose it to be n. 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 take a/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) is na/c : the difference between these is b/c , and therefore is known to you. For example, if a = 6 , b = 12 , c = 4 , and a/c = 11/2 , then the final result is 3.
Second Example/ (Bachet, problem xiii, p. 123: Bachet presented the above trick in a somewhat more general form, but one which is less effective in practice.) Ask A to take any number of counters that he pleases: suppose that he takes n counters. (i) Ask some one else, say B , to take p times as many, where p is any number you like to choose. (ii) Request A to give q of his counters to B , where q is any number you like to select. (iii) Next, ask B to transfer to A a number of counters equal to p times as many counters as A has in his possession. Then there will remain in B 's hands q (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 that B 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 has q (p + 1).
For example, if originally A took any number of counters, then (if you chose p equal to 2 ), first you would ask B to take twice as many counters as A had done; next (if you chose q equal to 3 ) you would ask A to give 3 counters to B ; and then you would ask B to give to A a number of counters equal to twice the number then in A 's possession; after this was done you would know that B had 3 (2 + 1) , that is, 9 left.
This trick (as also some of the following problems) may be performed equally well with one person, in which case A may stand for his right hand and B for his left hand.
Third Example Ask some one to perform in succession the following 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 the digits in this difference. (v) Add together the results of (iii) and (iv). Then the sum obtained as the result of this last operation will be 1089.
An illustration and the explanation of the rule are given below. 1
Fourth Example/ (Educational Times Reprints, 1890, vol. liii, p. 78.) The following trick depends on the same principle. 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) To reverse 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 difference 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.
For instance, take the sum 10. 17s. 5d.; we have
The following work explains the rule, and shows that the final result is independent of the sum written down initially.
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/ (Bachet, problem ix, p. 107.) 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 whether A selected the even or the odd number. Ask A to multiply his number by 2 (or any even number) and B to multiply his by 3 (or any odd number). Request them
to add the two products together and tell you the sum. If it is even, then originally A selected the odd number, but if it is odd, then originally A selected the even number. The reason is obvious.
Second Example/ (Bachet, problem xi, p. 113.) The above rule was extended by Bachet to any two numbers, provided they were prime to one another and one of them was not itself a prime. Let the numbers be m and n , and suppose that n is exactly divisible by p. Ask A to select one of these numbers, and B to take the other. Choose a number prime to p , say q. Ask A to multiply his number by q , and B to multiply his number by p. 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 by p. 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 × 10^3) + (0 × 10^2) + (1 × 10)+ 7 ; that is, the 2 represents 2 thousands, the product of 2 and 10^3 , the 0 represents 0 hundreds, the product of 0 and 10^2 ; the 1 represents 1 ten, the product of 1 and 10 , and the 7 represents 7 units. Thus every digit has a local value.
The application to tricks connected with numbers will be understood readily from three illustrative examples.
First Example/ (Some similar questions were given by Bachet in problem xii, p. 117; by Oughtred in his Mathematicall Recreations (translated from or founded on van Etten's work of 1633), London, 1653, problem xxxiv; and by Ozanam, part i, chapter x.) 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 by 5 , to add 7 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 numbers a 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.
Second Example (Bachet gave some similar questions in problem xii, p. 117.) 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) Add 3 to the product; the result is 2a + 3. (iii) Multiply this by 5 , and add 7 to the product; the result is 10a + 22. (iv) To this sum add the second number. (v) Multiply the result by 2. (vi) Add 3 to the product. (vii) Multiply by 5 , and add the third number to the product. The result is 100a+ 10b + c + 235. Hence, if the final result is known, it is sufficient to subtract 235 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 (A similar question was given by Laisant and Perrin in their Algèbre, Paris, 1892; and in L'Illustration for July 13, 1895.) 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 add 2 ; (iii) multiply the result by 2 ; (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. Let a 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 let a be the hundreds digit of the number chosen, b be the tens digit, and c be the units digit. Therefore the number is 100a + 10b +c. The number obtained by reversing the digits is 100c + 10b + a. The difference of these numbers is equal to (100a + c)-(100c + a) , that is, to 99 (a-c). But a-c is not greater than 9 , and therefore the remainder can only be 99 , 198 , 297 , 396 , 495 , 594 , 693 , 792 , or 891 --in each case the middle digit being 9 and the digit before it (if any) being equal to the difference between 9 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 attention 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 notations for fractions being allowed. One solution is 80.69 +.74 +.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 is 35/70 and 148/296. If the sum were 100 , a solution would be 50 , 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 is 100 ; a solution is 78 , 15 , sqrt(9) , and sqrt(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 illustrations 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 a watch, the last three are best exemplified by the use of a pack of playing cards, which readily lend themselves to such illustrations, and I present them in these forms.
First Example (Bachet, problem xx, p. 155; Oughtred, Mathematicall Recreations, London, 1653, p. 28.) 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 as m , (m + 1) , &c., the (n +12) th tap will be on the hour he thought of. For example, if he thinks of v and touches ix, then, if he taps successively ix, viii, vii, vi, ..., going backwards and reckoning them respectively as 5 , 6 , 7 , 8 , ..., the tap which he reckons as 21 will be on the v.
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 hour m : also it will make no difference if in addition he goes over 12 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 than n.
Second Example The following is another well-known way of indicating 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 to vi, v, &c., and if the person who selected the number counts the taps, reckoning from the hour selected (thus, if he selected x, he would reckon the first tap as the 11 th), then the 20 th tap as reckoned by him will be on the hour chosen.
For suppose he selected the n th hour. Then the 8 th tap is on xii and is reckoned by him as the (n + 8) th. The tap which he reckons as (n + 9) th is on xi, 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 as 20 th is on the hour n. Of course the hours indicated by the first seven taps are immaterial.
Extension It is obvious that the same trick can be
performed with any collection of m 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 n th thing is selected by a spectator. Then the first (19-m) taps are immaterial, the (20-m) th tap must be on the m th 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 the n th thing and is reckoned as the 20 th tap.
Third Example The following example rests on an chapter:1:cards 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 of n cards is given to some one who is asked to select one out of the first m cards and to remember (but not to mention) what is its number from the top of the pack (say it is actually the x th card in the pack). Then take the pack, reverse the order of the top m cards (which can be easily effected by shuffling), and transfer y 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 (A particular case of this problem was given by Bachet, problem xvii, p. 138.) 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 is r , then the sum of the number of pips on the bottom cards of all the piles will be 13 (p - 4) + r.
For, if x is the number of pips on the bottom card of a pile, the number of cards in that pile will be 13 - 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 the p piles and the r cards left over. Therefore (13 - x_1) + (13 - x_2) +... + (13 - x_p) + r = 52,; Therefore 13p - (x_1 + x_2 +... + x_p) + r = 52,; Therefore x_1 + x_2 +... + x_p = 13p - 52 + r; = 13 (p - 4) + r.
More generally, if a pack of n 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 to m , 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 (Bachet, problem xix, p. 152.). 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 upwards to some one, and asking him to select a card and state its number, reckoning from the top; suppose it to be the n th. Next, ask him to choose a number at which it shall appear in the pack; suppose he selects the m th. Take the pack and secretly move m - n cards from the bottom to the top (or if n is greater than m , then n - m from the top to the bottom) and of course the card will be in the required position.
Medieval Problems in Arithmetic.
Before leaving the subject of these elementary questions, I may mention a few problems which for centuries have appeared in nearly every collection of mathematical recreations, and therefore may claim what is almost a prescriptive right to a place here.
First Example (Some similar problems were given by Bachet, appendix, problem iii, p. 206; problem ix, p. 233; by Oughtred in his Recreations, p. 22: and by Ozanam, 1803 edition, vol. i, p. 174; 1840 edition, p. 79. Earlier instances occur in Tartaglia's writings.) The following is a sample of one class of these puzzles. Three men robbed a gentleman of a vase, containing 24 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. Vase Problem|)
Second Example (Bachet, problem xxii, p. 170.) 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, and so on. He wins who is the first to reach (say) 50. Obviously, if A calls 43 , then whatever B adds to that, A can win next time. Similarly, if A calls 36 , B cannot prevent A 's calling 43 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 than m may be added at any one time, and n is the number to be called by the victor, then the key numbers 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 placing n coins, matches, or other objects on a table, and directing each player in turn to take away not more than m of them. Whoever takes away the last coin wins. Obviously the key numbers are multiples of m + 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 of m + 1.
Mr Loyd has also suggested (Tit-Bits, London, July 17, Aug. 7, 1897.) 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 than m of them which are in unbroken sequence: m being less than
n 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; then B 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 to 50 , but he loses if he plays a card which makes this sum exceed 50. The game can be played mentally or by noting the numbers on a piece of paper, and in practice it is unnecessary to use cards.
Thus, if they play as follows A , 4 ; B , 3 ; A , 1 ; B , 6 ; A , 3 ; B , 4 ; A , 4 ; B , 5 ; A , 4 ; B , 4 ; A , 5 ; the game stands at 43. B can now win, for he may safely play 3 , since A has not another 4 wherewith to follow it; and if A plays less than 4 , B will win the next time. Again, if they play 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 play 1 , and B wins by playing 1.
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 (Bachet, problem xxv, p. 187.).
Place 24 counters on a table. Ask P to take one counter,
Q to take two counters, and R to take three counters. Next, ask the person who selected a to take as many counters as he has already, whoever selected e to take twice as many counters as he has already, and whoever selected 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 line Par 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 by P , the vowel in the second syllable indicates the thing selected by Q , and of course R selected the remaining thing. Salve certa animae semita vita quies was suggested by Oughtred (Mathematicall Recreations, London, 1653, p. 20.) as an alternative mnemonic line.
Extension M. Bourlet, in the course of a very kindly notice (Bulletin des sciences mathématiques, Paris, 1893, vol. xvii, pp. 105--107.) 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 of n people, P_0, P_1, P_2..., P_(n-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 numbers 0, 1..., n-1 , instead of by vowels. Give one counter to P_1 , two counters to P_2 , and generally k counters to P_k. Note the number of counters left on the table. Next ask the person who had chosen the domino 0 to take as many counters as he had already, and generally whoever had chosen the domino h to take n^h times as many dominoes as he had already: thus if P_k had chosen the domino numbered h , he
would take n^h k counters. Note the total number of counters taken, n^hk. Divide it by n , then the remainder will be the number on the domino selected by P_0 ; divide the quotient by n , and the remainder will be the number on the domino selected by P_1 ; divide this quotient by n , and the remainder will be the number on the domino selected by P_2 ; and so on. In other words, if the number of counters taken is expressed in the scale of notation whose radix is n , then the (h + 1) th digit from the right will give the number on the domino selected by P_h.
Thus in Bachet's problem with 3 people and 3 dominoes, we should first give one counter to Q , and two counters to R , 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 marked 1 or e to take three times as many counters as he had already, and whoever selected the domino marked 2 or i to take nine times as many counters as he had already. By noticing the original number of counters, and observing that 3 of these had been given to Q and R , we should know the total number taken by P , Q , and R. If this number were divided by 3 , the remainder would be the number of the domino chosen by P ; if the quotient were divided by 3 the remainder would be the number of the domino chosen by Q ; and the final quotient would be the number of the domino chosen by R.
I may add that Bachet also discussed the case when n = 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 The last of these antique problems to which I referred consists in placing men round a circle so that if every n th 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 (De Bello Judaico, bk. iii, chaps. 16--18.) says that Josephus saved his life by such a device. According to his account, after the Romans had captured Jotopat, Josephus 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 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 suggested 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 the 31 st and 16 th 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 that every ninth man should be cast overboard, reckoning from a certain point. It is desired to find an arrangement by which all the Christians should be saved. (Bachet, problem xxiii, p. 174. The same problem had been previously enunciated by Tartaglia.)
Problems like this can be easily solved by counting, but it is impossible to give a general rule. In this case, the Christians, reckoning from the man first counted, must occupy the places 1 , 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 is
4 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.
I insert next some instances
of demonstrations (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's Algebra, 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 that 1 = 2 was given by R. Chartres in Knowledge, 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 to y (from y = 0 to y = 1 ) of the integral to x (from x=0 to x = 1 ) of φ , but if φ=(x^2-y^2)/(x^2 + y^2)^2 , this gives 1/4π =-1/4π .) leading to arithmetical results which are obviously 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 interesting specimen--is as follows. Suppose that a = b , then ab = a^2. ; Therefore ab-b^2 = a^2 - b^2. ; Therefore b(a-b) = (a+b)(a-b). ; Therefore b = a + b. ; Therefore b = 2b. ; Therefore 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.; Therefore (a + b)(a - b) = 2c(a - b).; Therefore a^2 - 2ac = b^2 - 2bc.; Therefore a^2 - 2ac + c^2 = b^2 - 2bc + c^2.; Therefore (a - c)^2 = (b - c)^2.; Therefore a = b.
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, Therefore 2 (-1) = 1 = 0. ; Therefore (-1) = 0. ; Therefore -1 = e^0. ; Therefore -1 = 1.;
The same argument may be expressed thus. Let x be a quantity which satisfies the equation e^x = -1.; Square both sides, Therefore e^2x = 1. ; Therefore 2x = 0.; Therefore x = 0.; Therefore e^x = e^0.; But $e^x = -1$ and $e^0 = 1$, Therefore -1 = 1.;
Fourth Fallacy As yet another instance, we know that (1 + x) = x - 1/2x^2 + 1/3x^3 -.... If x = 1 , the resulting series is convergent; hence we have 2 = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8 + 1/9 -.... ; Therefore 2 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 2 = 1 + 1/3 - 1/2 + 1/5 + 1/7 - 1/4 + 1/9-... ; = 1 - 1/2 + 1/3 - 1/4 + 1/5-... ; = 2. ; Hence 2 = 1.;
Fifth Fallacy This fallacy is very similar to that last given. We have 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 +...} - (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 sqrt(-1) = sqrt(-1) in the form
sqrt(-1/1) = sqrt(1/-1), ; hence sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1), ; therefore (sqrt(-1))^2 = (sqrt(1))^2, ; that is, -1 = 1.
Seventh Fallacy Again, we have sqrt(a) × sqrt(b) = sqrt(ab). ; Hence sqrt(-1) × sqrt(-1) = sqrt((-1) (-1)), ; therefore (sqrt(-1))^2 =sqrt(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, sqrt(x-y) = i sqrt(y-x) i where i stands either for + sqrt(-1) or for - sqrt(-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 , Therefore sqrt(a - b) = i sqrt(b - a) ii Next put x = b and y = a , Therefore sqrt(b - a) = i sqrt(a - b) iii Also since (i) is an identity, it follows that in (ii) and (iii) the symbol i must be the same, that is, it represents + sqrt(-1) or - sqrt(-1) in both cases. Hence, from (ii) and (iii), we have sqrt(a-b); sqrt(b - a) = i^2 sqrt(b-a); sqrt(a - b), ; Therefore 1 = i^2, ; that is 1 = -1.
Ninth Fallacy The following fallacy is due to D'Alembert (Opuscules mathématiques, Paris, 1761, vol. i, p. 201.). 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 proportion a > b , then c > d. Now if we put a = d =1 and b = c = -1 we have four numbers which satisfy the relation ad = bc and such that a>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 (See Nature, Feb. 15, March 1, 1894, vol. xlix, pp. 365--366, 413.) 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, is 1/8 ; similarly the probability that all three come down tail is 1/8 : hence the probability that all the coins come down alike ( either all of them heads or all of them tails) is 1/4. But, of three coins thus thrown up, at least two must come down alike; now the probability that the third coin comes down head is 1/2 and the probability that it comes down tail is 1/2 , 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 following questions, which I have often propounded in past years: though not fallacies, they may serve to illustrate the fact that the answer to 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 bets 1/n th of his money on an even chance (say tossing heads or tails with a penny): he repeats this again and again, each time betting 1/n th 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 generalised as follows (The substance of the problem was given in a scholarship paper set at Cambridge about 30 years ago, and possibly was not new then.). Construct a rectangular board of mn cells (or small squares) by ruling m+1 vertical lines and n+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 containing 16 cells ( one-quarter of a chess-board), where m=n = 4 , there are 70 such routes; while on a common chess-board, where m=n=8 , there are no less than 12870 such routes. A similar theorem can be enunciated for a parallelopiped.
Another problem of a somewhat similar type is the determination of the number of closed routes through mn points arranged in m rows and n columns, following the lines of the quadrilateral net-work, and passing once and only once through each point (See C.F. Sainte-Marie in L'Intermédiaire des mathématiciens, Paris, vol. xi, March, 1904, pp. 86--88.).
Permutation Problems. Permutation Problems As other simple illustrations of
the very large number of ways in which combinations of even a few things can be arranged, I may note that as many as 19,958400 distinct skeleton cubes can be formed with twelve differently coloured rods of equal length (Mathematical Tripos, Cambridge, Part I, 1894.); again there are 3,979614,965760 ways of arranging a set of twenty-eight dominoes ( a set from double zero to double six) in a line, with like numbers in contact (Reiss in Annali di matematica, Milan, November, 1871, vol. v, pp. 63--120.); while there are no less than 53644,737765,488792,839237,440000 possible different distributions of hands at whist with a pack of fifty-two cards (That is (52!)/(13!)^4 .).
Voting Problems Here is a simple example on combinations dealing with the cumulative vote as affecting the representation of a minority. If there are p electors each having r votes of which not more than s may be given to one candidate, and n men are to be elected, then the least number of supporters who can secure the election of a candidate must exceed pr/(ns + r).
Exploration Problems Another common question is concerned with the maximum distance into a desert which could be reached from a frontier settlement by the aid of a party of n explorers, each capable of carrying provisions that would last one man for a days. The answer is that the man who reaches the greatest distance will occupy na/(n + 1) days before he returns to his starting point. If in the course of their journey they may make depôts, the longest possible journey will occupy 1/2a (1 + 1/2 + 1/3+... + 1/n) days. Further extensions by the use of horses and cycles will suggest themselves.
Here I conclude my account of such of these easy problems on numbers or elementary algebra as seemed worth reproducing. It will be noticed that the majority of them either are due to Bachet or were collected by him in his classical Problèmes; but it should be added that besides the questions I have mentioned
he enunciated, even if he did not always solve, some other problems of greater interest. One instance will suffice.
[Bachet's Weights Problem]Bachet's Weights Problem (Bachet, Appendix, problem v, p. 215.) Among the more difficult problems proposed by Bachet was the determination of the least number of weights which would serve to weigh any integral number of pounds from 1 lb. to 40 lbs. inclusive. Bachet gave two solutions: namely, (i) the series of weights of 1 , 2 , 4 , 8 , 16 , and 32 lbs.; (ii) the series of weights of 1 , 3 , 9 , and 27 lbs.
If the weights may be placed in only one of the scale-pans, the first series gives a solution, as had been pointed out in 1556 by Tartaglia (Trattato de' numeri e misure, Venice, 1556, vol. ii, bk. i, chap. xvi, art. 32.).
Bachet, however, assumed that any weight might be placed in either of the scale-pans. In this case the second series gives the least possible number of weights required. His reasoning is as follows. To weigh 1 lb. we must have a 1 lb. weight. To weigh 2 lbs. we must have in addition either a 2 lb. weight or a 3 lb. weight; but, if we are confined to only one new weight (in addition to the 1 lb. we have got already), then with no weight greater than 3 lbs. could we weigh 2 lbs.: if we use a 2 lb. weight we then can weigh 1 lb., 2 lbs., and 3 lbs., but if we use a 3 lb. weight we then can weigh 1 lb., (3-1) lbs., 3 lbs., and (3 + 1) lbs.; hence a 3 lb. weight is preferable. Similarly, to enable us to weigh 5 lbs. we must have another weight not greater than 9 lbs., and a weight of 9 lbs. enables us to weigh every weight from 1 lb. to 13 lbs.; hence it is the best to choose. The next weight required will be 2(1 + 3 + 9) + 1 lb., that is, will be 27 lbs.; and this enables us to weigh from 1 lb. to 40 lbs. Thus only four weights are required, namely, 1 lb., 3 lbs., 3^2 lbs., and 3^3 lbs.
We can show similarly that the series of weights of 1 , 3 , 3^2 , ... , 3^(n-1) lbs. will enable us to weigh any integral number of pounds from 1 lb. to (1+3 + 3^2+... 3^(n-1)) lbs., that is, to
1/2(3^n-1) lbs. This is the least number with which the problem can be effected.
To determine the arrangement of the weights to weigh any given mass we have only to express the number of pounds in it as a number in the ternary scale of notation, except that in finding the successive digits we must make every remainder either 0 , 1 , or -1 : to effect this a remainder 2 must be written as 3-1 , that is, the quotient must be increased by unity, in which case the remainder is -1. This is explained in most text-books on algebra.
Bachet's argument does not prove that his result is unique or that it gives the least possible number of weights required. These omissions have been supplied by Major MacMahon, who has discussed the far more difficult problem (of which Bachet's is a particular case) of the determination of all possible sets of weights, not necessarily unequal, which enable us to weigh any integral number of pounds from 1 to n inclusive, (i) when the weights may be placed in only one scale-pan, and (ii) when any weight may be placed in either scale-pan. He has investigated also the modifications of the results which are necessary when we impose either or both of the further conditions (a) that no other weighings are to be possible, and (b) that each weighing is to be possible in only one way, that is, is to be unique (See his article in the Quarterly Journal of Mathematics, 1886, vol. xxi, pp. 367--373. An account of the method is given in Nature, Dec. 4, 1890, vol. xlii, pp. 113--114.).
The method for case (i) consists in resolving 1 + x + x^2 +... + x^n into factors, each factor being of the form 1 + x^a+ x^2a +... +x^ma ; the number of solutions depends on the composite character of n + 1. The method for case (ii) consists in resolving the expression x^-n + x^(-n+1)+... + x^-1 + 1 + x +... + x^(n-1) + x^n into factors, each factor being of the form x^(-ma) +... + x^-a + 1 + x^a+... + x^ma ; the number of solutions depends on the composite character of 2n+1.
Bachet's problem falls under case (ii), n = 40. MacMahon's
analysis shows that there are eight such ways of factorizing x^(-40) + x^(-39) +...+ 1 + x^39 + x^40. First, there is the expression itself in which a=1 , m= 40. Second, the expression is equal to (1-x^81)/x^40 (1-x) , which can be resolved into the product of (1-x^3)/x (1-x) and (1-x^81)/x^39 (1-x^3) ; hence it can be resolved into two factors of the form given above, in one of which a=1 , m=1 , and in the other, a =3 , m=13. Third, similarly, it can be resolved into two such factors, in one of which a=1 , m = 4 , and in the other a = 9 , m = 4. Fourth, it can be resolved into three such factors, in one of which a = 1 , m= 1 , in another a = 3 , m = 1 , and in the other, a = 9 , m = 4. Fifth, it can be resolved into two such factors, in one of which a = 1 , m = 13 , and in the other a = 27 , m = 1. Sixth, it can be resolved into three such factors, in one of which a=1 , m = 1 , in another a = 3 , m = 4 , and in the other a = 27 , m = 1. Seventh, it can be resolved into three such factors, in one of which a = 1 , m= 4 , in another a = 9 , m=1 , and in the other a= 27 , m=1. Eighth, it can be resolved into four such factors, in one of which a= 1 , m= 1 , in another a = 3 , m = 1 , in another a = 9 , m = 1 , and in the other a = 27 , m= 1.
These results show that there are eight possible sets of weights with which any integral number of pounds from 1 to 40 can be weighed subject to the conditions (ii), (a), and (b). If we denote p weights each equal to w by w^p , these eight solutions are 1^40 ; 1 , 3^13 ; 1^4 , 9^4 ; 1 , 3 , 9^4 ; 1^13 , 27 ; 1 , 3^4 , 27 ; 1^4 , 9 , 27 ; 1 , 3 , 9 , 27. The last of these is Bachet's solution: not only is it that in which the least number of weights are employed, but it is also the only unique one in which all the weights are unequal
.
At the commencement
of this chapter I alluded to the special interest which many mathematicians find in the theorems of higher arithmetic: such, for example, as that every prime of the form 4n+1 and every power of it is expressible as the sum of two
squares (Fermat's Diophantus, Toulouse, 1670, bk. iii, prop. 22, p. 127; or Brassinne's Précis, Paris, 1853, p. 65.), and the first and second powers can be expressed thus in only one way. For instance, 13 = 3^2 + 2^2 , 13^2 = 12^2 + 5^2 , 13^3 = 46^2 + 9^2 , and so on. Similarly 41 = 5^2 + 4^2 , 41^2 = 40^2 + 9^2 , 41^3 = 236^2+115^2 , and so on.
Propositions such as the one just quoted may be found in text-books on the theory of numbers and therefore lie outside the limits of this work, but there are one or two questions in higher arithmetic involving points not yet quite cleared up which may find a place here.
Primes The first of these is concerned with the possibility of determining readily whether a given number is prime or not. Euler and Gauss attached great importance to this problem, but failed to establish any conclusive test. It would seem, however, that Fermat possessed some means of finding from its form whether a given number (at any rate if one of certain known forms) was prime or not. Thus, in answer to Mersenne who asked if he could tell without much trouble whether the number 100895,598169 was a prime, Fermat wrote on April 7, 1643, that it was the product of 898423 and 112303 , both of which were primes. I have indicated elsewhere one way by which this result can be found, and Mr F.W. Laurence has indicated another which may have been that used by Fermat in this particular case.
Mersenne's Numbers. Mersenne's Numbers (For references, see chapter ix below.) Another illustration, confirmatoryMersenne:I of the opinion that Fermat or some of his contemporaries had a test by which it was possible to find out whether certain numbers were prime, may be drawn from Mersenne's Cogitata Physico-Mathematica which was published in 1644. In the preface to that work it is asserted that in order that 2^p-1 may be prime, the only values of p , not greater than 257 , which are possible are 1 , 2 , 3 , 5 , 7 , 13 , 17 , 19 , 31 , 67 , 127 , and 257 : I conjecture that the number 67 is a misprint for 61. With this correction the statement appears to be
trueIt's not: see http://www.mersenne.orgwww.mersenne.org, and it has been verified for all except nineteen values of p : namely, 71 , 101 , 103 , 107 , 109 , 137 , 139 , 149 , 157 , 163 , 167 , 173 , 181 , 193 , 199 , 227 , 229 , 241 , and 257. Of these values, Mersenne asserted that p = 257 makes 2^p-1 a prime, and that the other values make 2^p-1 a composite number. The demonstrations for the cases when p = 89 , 127 have not been published; nor have the actual factors of 2^p-1 when p = 89 been as yet determined: the discovery of these factors may be commended to those interested in the theory of numbers.
Mersenne's result could not be obtained empirically, and it is impossible to suppose that it was worked out for every case; hence it would seem that whoever first enunciated it was acquainted with certain theorems in higher arithmetic which have not been re-discovered.
Perfect Numbers. Perfect Numbers (On the theory of perfect numbers, see bibliographical references by H. Brocard, L'Intermédiaire des mathématiciens, Paris, 1895, vol. ii, pp. 52--54; and 1905, vol. xii, p. 19.) The theory of perfect numbers depends directly on that of Mersenne's Numbers. A number is said to be perfect if it is equal to the sum of all its integral subdivisors. Thus the subdivisors of 6 are 1 , 2 , and 3 ; the sum of these is equal to 6 ; hence 6 is a perfect number.
It is probable that all perfect numbers are included in the formula 2^(p-1) (2^p-1) , where 2^p-1 is a prime. Euclid proved that any number of this form is perfect; Euler showed that the formula includes all even perfect numbers; and there is reason to believe--though a rigid demonstration is wanting--that an odd number cannot be perfect. If we assume that the last of these statements is true, then every perfect number is of the above form. It is easy to establish that every number included in this formula (except when p = 2 ) is congruent to unity to the modulus 9 , that is, when divided by 9 leaves a remainder 1 ; also that either the last digit is a 6 or the last two digits are 28.
Thus, if p = 2 , 3 , 5 , 7 , 13 , 17 , 19 , 31 , 61 , then by Mersenne's rule
the corresponding values of 2^p-1 are prime; they are 3 , 7 , 31 , 127 , 8191 , 131071 , 524287 , 2147483647 , 2305843009213693951 ; and the corresponding perfect numbers are 6 , 28 , 496 , 8128 , 33550336 , 8589869056 , 137438691328 , 2305843008139952128 , and 2658455991569831744654692615953842176.
Goldbach's Theorem. Goldbach's Theorem Another interesting problem in higher arithmetic is the question whether there are any even integers which cannot be expressed as a sum of two primes. Probably there are none. The expression of all even'even' inserted as per errata sheet integers not greater than 5000 in the form of a sum of two primes has been effected (Transactions of the Halle Academy (Naturforschung), vol. lxxii, Halle, 1897, pp. 5--214: see also L'Intermédiaire des mathématiciens, 1903, vol. x, and 1904, vol. xi.), but a general demonstration that all even integers can be so expressed is wanting.
Lagrange's Theorem. Lagrange's Theorem (Nouveaux Mémoires de l'Académie Royale des Sciences, Berlin, 1775, p. 356.) Another theorem in higher arithmetic which, as far as I know, is still unsolved, is to the effect that every prime of the form 4n - 1 is the sum of a prime of the form 4n + 1 and of double a prime of the form 4n + 1 ; for example, 23 = 13 + 2 × 5. Lagrange, however, added that it was only by induction that he arrived at the result.
Fermat's Theorem on Binary Powers tocsubsectionFermat's Theorem on Binary Powers Fermat's Theorem on Binary Powers. Fermat
enriched mathematics with a multitude of new propositions. With two exceptions all these have been proved subsequently to be true. The first of these exceptions is his theorem on binary powers, in which he asserted that all numbers of the form 2^m + 1 , where m = 2^n , are primes (Letter of Oct. 18, 1640, Opera, Toulouse, 1679, p. 162: or Brassinne's Précis, p. 143.), but he added that, though he was convinced of the truth of this proposition, he could not obtain a valid demonstration.
It may be shown that 2^m +1 is composite if m is not a power of 2 , but of course it does not follow that 2^m + 1 is a prime if m is a power of 2. As a matter of fact the theorem
is not true. In 1732 Euler (Commentarii Academiae Scientiarum Petropolitanae, St Petersburg, 1738, vol. vi, p. 104; see also Novi Comm. Acad. Sci. Petrop., St Petersburg, 1764, vol. ix, p. 101: or Commentationes Arithmeticae Collectae, St Petersburg, 1849, vol. i, pp. 2, 357.) showed that if n = 5 the formula gives 4294,967297 , which is equal to 641 × 6,700417 : curiously enough, these factors can be deduced at once from Fermat's remark on the possible factors of numbers of the form 2^m ± 1 , from which it may be shown that the prime factors (if any) of 2^32 + 1 must be primes of the form 64n + 1.
During the last thirty years it has been shown (For the factors and bibliographical references, see the memoir by A.J.C. Cunningham and A.E. Western, Transactions of the London Mathematical Society, May 14, 1903, series 2, vol. i, p. 175.) that the resulting numbers are composite when n = 6 , 9 , 11 , 12 , 18 , 23 , 36 , and 38 : the two last numbers contain many thousands of millions of digits. I believe that Eisenstein asserted that the number of primes of the form 2^m + 1 , where m = 2^n , is infinite: the proof has not been published, but perhaps it might throw some light on the general theory.
Fermat's Last Theorem tocsubsectionFermat's Last Theorem Fermat's Last Theorem. I pass now to the only other assertion made by Fermat which has not been proved hithertoAndrew Wiles' proof appeared in 1995: Annals of Mathematics, vol. cxli, pp. 443--551. This, which is sometimes known as Fermat's Last Theorem, is to the effect (Fermat's enunciation will be found in his edition of Diophantus, Toulouse, 1670, bk. ii, qu. 8, p. 61; or Brassinne's Précis, Paris, 1853, p. 53. For bibliographical references, see L'Intermédiaire des mathématiciens, 1905, vol. xii, pp. 11, 12.) that no integral values of x , y , z can be found to satisfy the equation x^n + y^n = z^n , if n is an integer greater than 2. This proposition has acquired extraordinary celebrity from the fact that no general demonstration of it has been given, but there is no reason to doubt that it is true.
Fermat seems to have discovered its truth first (See a letter from Fermat quoted in my History of Mathematics, London, chapter xv.) for the case n = 3 , and then for the case n = 4. His proof for the former of these cases is lost, but that for the latter is extant (Fermat's Diophantus, note on p. 339; or Brassinne's Précis, p. 127.), and a
similar proof for the case of n=3 was given by Euler (Euler's Algebra (English trans. 1797), vol. ii, chap. xv, p. 247.). These proofs depend upon showing that, if three integral values of x , y , z can be found which satisfy the equation, then it will be possible to find three other and smaller integers which also satisfy it: in this way finally we show that the equation must be satisfied by three values which obviously do not satisfy it. Thus no integral solution is possible. It would seem that this method is inapplicable except when n = 3 and n = 4.
Fermat's discovery of the general theorem was made later. An easy demonstration can be given on the assumption that every number can be resolved into prime (complex) factors in one and only one way. That assumption has been made by some writers, but it is not universally true. It is possible that Fermat made some such supposition, though it is perhaps more probable that he discovered a rigorous demonstration. At any rate he asserts definitely that he had a valid proof--demonstratio mirabilis sane--and the fact that every other theorem on the subject which he stated he had proved has been subsequently verified must weigh strongly in his favour; especially as in making the one statement in his writings which is not correct he was scrupulously careful to add that he could not obtain a satisfactory demonstration of it.
It must be remembered that Fermat was a mathematician of quite the first rank who had made a special study of the theory of numbers. That subject is in itself one of peculiar interest and elegance, but its conclusions have little practical importance, and since his time it has been discussed by only a few mathematicians, while even of them not many have made it their chief study. This is the explanation of the fact that it took more than a century before some of the simpler results which Fermat had enunciated were proved, and thus it is not surprising that a proof of the theorem which he succeeded in establishing only towards the close of his life should involve great difficulties.
In 1823 Legendre (Reprinted in his Théorie des Nombres, Paris, 1830, vol. ii, pp. 361--368: see also pp. 5, 6.) obtained a proof for the case of n = 5 ; in 1832 Lejeune Dirichlet (Crelle's Journal, 1832, vol. ix, pp. 390--393.) gave one for n=14 , and in 1840 Lamé and Lebesgue (Liouville's Journal, 1841, vol. v, pp. 195--215, 276--9, 348--9.) gave proofs for n=7.
The proposition appears to be true universally, and in 1849 Kummer (References to Kummer's Memoirs are given in Smith's Report to the British Association on the Theory of Numbers, London, 1860.), by means of ideal primes, proved it to be so for all numbers except those (if any) which satisfy three conditions. It is not known whether any number can be found to satisfy these conditions, but it seems unlikely, and it has been shown that there is no number less than 100 which does so. The proof is complicated and difficult, and there can be little doubt is based on considerations unknown to Fermat. I may add that to prove the truth of the proposition when n is greater than 4 , it obviously is sufficient to confine ourselves to cases where n is a prime, and the first step in Kummer's demonstration is to show that in such cases one of the numbers x , y , z must be divisible by n.
Naturally there has been much speculation as to how Fermat arrived at the result. The modern treatment of higher arithmetic is founded on the special notation and processes introduced by Gauss, who pointed out that the theory of discrete magnitude is essentially different from that of continuous magnitude, but until the end of the last century the theory of numbers was treated as a branch of algebra, and such proofs by Fermat as are extant involve nothing more than elementary geometry and algebra, and indeed some of his arguments do not involve any symbols. This has led some writers to think that Fermat used none but elementary algebraic methods. This may be so, but the following remark, which I believe is not generally known, rather points to the opposite conclusion. He had proposed, as a problem to the
English mathematicians, to show that there was only one integral solution of the equation x^2 + 2 = y^3 : the solution evidently being x = 5, y = 3. On this he has a note (Fermat's Diophantus, bk. vi, prop. 19, p. 320; or Brassinne's Précis, p. 122.) to the effect that there was no difficulty in finding a solution in rational fractions, but that he had discovered an entirely new method--sane pulcherrima et subtilissima--which enabled him to solve such questions in integers. It was his intention to write a work (Fermat's Diophantus, bk. iv, prop. 31, p. 181; or Brassinne's Précis, p. 82.) on his researches in the theory of numbers, but it was never completed, and we know but little of his methods of analysis. I venture however to add my private suspicion that continued fractions played a not unimportant part in his researches, and as strengthening this conjecture I may note that some of his more recondite results--such as the theorem that a prime of the form 4n + 1 is expressible as the sum of two squares--may be established with comparative ease by properties of such fractions
.
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