CII. l] ARITHMETICAL RECREATIONS 23

he would take n^k counters. The total number of counters taken is '^n^k. Divide this by n, then the remainder will be the number on the domino selected by Pq ; divide the quotient by ??, and the remainder will be the number on the domino selected by Pi ; divide this quotient by n, and the remainder will be the number on the domino selected by P^ ; 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-\- l)th digit from the right will give the number on the domino selected byP/^.
Thus in Bachet's problem with 3 people and 3 dominoes, we should first give one counter to Q, and two counters to P, while P would have no counters ; then we should ask the person who had selected the domino marked 0 or a to take as many counters as he had already, whoever had selected the domino marked 1 or e to take three times as many counters as he had already, and whoever had 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 P, we should know the total number taken by P, Q, and P. 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 re- mainder would be the number of the domino chosen by Q; and the final quotient would be the number of the domino chosen by P.
Eojj^loration Problems. Another common question is con- cerned 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 nal{n + 1) days before he returns to his starting point. If in the course of their journey they may make depots, the longest possible journey will occupy ^ a (1 + i + J + . . . + Ijn ) days.
TJie Josephiis Problem. Another of these antique problems consists in placing men round a circle so that if every mth man is killed, the remainder shall be certain specified individuals. Such problems can be ea«ily tsolvcd ompiricaliy.
24 ARITHMETICAL RECREATIONS [CH. I
Hegesippus* says that Josephus saved his life by such a device. According to his account, after the Romans had captured Jotapat, 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 them- selves, 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 31st and 16th place respectively.
The medieval question was usually presented in the following form. A ship, carrying as passengers 15 Turks and 15 Christians, encountered a storm, and, in order to save the ship and crew, one-half of the passengers had to be thrown into the sea. Accordingly the passengers were placed in a circle, and every ninth man, reckoning from a certain point, was cast overboard. It is desired to find an arrangement by which all the Christians should be saved "f*. In this case we must arrange the men thus : GGGGTTTTTGCTGGGTCTTGGTTTGTTGGT, where G stands for a Christian and T for a Turk. The order can be recollected by the positions of the vowels in the follow- ing line : From numbers aid and art, never will fame depart, where a stands for l,e for 2, i for 3, o for 4, and u for 5. Hence the order is o Christians, u Turks, &c.
If every tenth man were cast overboard, a similar mnemonic line is Rex paphi cum gente bona dat signa serena. An oriental setting of this decimation problem runs somewhat as follows. Once upon a time, there lived a rich farmer who had 30 children, 15 by his first wife who was dead, and 15 by his second wife. The latter woman was eager that her eldest son should inherit the property. Accordingly one dny she said to him, " Dear Husband,
* De Bello Judaico, bk. in, chaps. 16 — 18.
t Bachet, problem xxiii, p, 174. The same problem had been previously enunciated by Tavtaglia.
CH. l] ARITHMETICAL RECREATIONS 25
you are getting old. We ought to settle who shall be your heir. Let us arrange our 30 children in a circle, and counting from one of them remove every tenth child until there remains but one, who shall succeed to your estate." The proposal seemed reasonable. As the process of selection went on, the farmer grew more and more astonished as he noticed that the first 14 to disappear were children by his first ^vife, and he ob- served that the next to go would be the last remaining member of that family. So he suggested that they should see what would happen if they began to count backwards from this lad. She, forced to make an immediate decision, and reflecting that the odds were now 15 to 1 in favour of her family, readily assented. Who became the heir ?
In the general case n men are arranged in a circle which is closed up as individuals are picked out. Beginning anywhere, we continually go round, picking out each mth man until only r are left. Let one of these be the man who originally occu- pied the pth place. Then had we begun with n+1 men, he would have originally occupied the (p-f-m)th place when p-{-m is not greater than n+1, and the (p + m. — n — l)th. place when p + m is greater than n + 1. Thus, provided there are to be r men left, their original positions are each shifted forwards along the circle ?n places for each addition of a single man to the original group*.
Now suppose that with n men the last survivor (r = 1) occupied originally the p\h place, and that with {n-\-x) men the last survivor occupied the yih. place. Then, if we confine ourselves to the lowest value of x which makes y less than m, we have y=(p-\- mx) —(n + x).
Based on this theorem we can, for any specified value of n, calculate rapidly the position occupied by the last survivor of the company. In effect, Tait found the values of n for which a man occupying a given position p, which is less than m, would be the last survivor, and then by repeated applications of the proposition, obtained the position of the survivor for intermediate values of n. * P. G. Tait, Collected Scientljic Fa^en, Cambridge, vol. n. 1900, pp. 432—435.
2g ARITHMETICAL RECREATIONS [CH. I
For instance, take the Josephus V^o^^^.t ^ToZv^d ThPn we know that the final survivor of 41 men o Then we know ^^_^^ ^^^^ ^^^^^ ^^ T^^en
:7:7Z th t.^vor oL'upied originally the .th place.
ite:lr;onsider only the --J^ ^tl^I ^^^^^^^^^^^
, less than «' /^ ^^^X/^/f^tiJ^ X , ^sitive and Now, we have to take a value ot x w ii ^ r
" = \^.;Sr;S 349956 52i934, or 787401; and the 69127 If 5536 233304^ 3 14 21 47, 158, 237, 533, 1199, 4046, second place when »- J'f'^*' t'' ' nn65S. From these cAflQ i^fi'i'5 46085, 103691, 1181 lO^.or ill iujt'- ...
lXhr.;pe-dapplicat.ns^^^
any intermediate values of « ^^^ P"— f,^ go^lth place ;
by the man last taken. JJ^^ ^'^fj" ^^^ ^ ^h 1000000 men,
with 100000 men, the 92620th place ana
.V, R'^TTQSth Dlace are those which would be seleccea oy
the 637798tn puce aie „„hiected to tnmation.
bimiiaiiy
!='=,'^ « ?.t?° «? C« r- ,tt;'"^
CH. l]
ARITHMETICAL RECREATIONS
27
every Kh man is selected, all the Christians will be picked out for punishment. The problem is to find a, 6, h, and k.
I suggest as a similar problem, to find an arrangement of c Turks and c Christians arranged in a circle, so that if beginning at a particular man, say the first, every hth. man is selected, all the Turks will be picked out, but if, beginning at the same man, every A;th man is selected, all the Christians will be picked out. This makes an interesting question because it is conceivable that the operator who picked out the victims might get confused and take h instead of h, or vice versa, and so consign all his iriends to execution instead of those whom he had intended to pick out. The problem is, for any given value of c, to find an arrangement of the men and the corresponding suitable values of h and h. Obviously if c = 2, then for an arrangement like T C G T Si solution is A = 4, A; = 3. If c = 3, then for an arrange- ment like TO TOOT a solution is /t = 7, A; = 8. If c = 4, then for an arrangement like T G T T G T G C a solution is h = 9, A; = 5 ; and so on. Is it possible to give similar arrangements for higher numbers ?
ADDENDUM.
Note. Page 13. Solutions of the ten digit problems are 35/70 + 148/296 = 1, or -01234 + -98765 = 1; and 50 + 49 + 1/2 + 38/76 = 100.
A solution of the nine digit problem is
1-234 + 98-765=100, or 97 + 8/12 + 4/6 + 5/3 = 100; but if an algebraic sum is permissible a neater solution is
123-45-67 + 89 = 100, where the digits occur in their natural order.
Note. Page 18. There are several solutions of the division of 24 ounces under the conditions specified. One of these solutions is as follows:
The vessels can contain 24 oz. 13 oz. 11 oz. 5 oz.
Their contents originally are... 24... 0... 0... 0...
First, make their contents 0... 8... 11... 5...
Second, make their contents ... 16... 8... 0... 0... Third, make their contents ... 16 ... 0 ... 8 ... 0 ... Fourth, make their contents ... 3 ... 13 ... 8 ... 0 ...
Fifth, make their contents 3 ... 8 ... 8 ... 5 ...
Lastly, make their contents ... 8... 8... 8... 0...
Note. Pages 26 — 27. The simplpst solution of the five Christians and five Turks problem isa=l, /i = ll, 6 = 9, A = 29.
28