CH. i] ARITHMETICAL RECREATIONS 27 



every kth man is selected, all the Christians will be picked out 

 for punishment. The problem is to find a, b, h, and k. A solution 

 isa = l, A. = ll,6 = 9, k = 29. 



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 kth. 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 k instead of h, or vice versa, and so consign all his 

 friends 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 k. Obviously if c = 2, then for an arrangement like 

 T C C T a solution is h = 4, k = 3. If c = 3, then for an arrange- 

 ment like TGTGGT a solution is h = 7, k = 8. If c = 4, then 

 for an arrangement like TGTTGTGG a solution is h = 9, 

 k = 5. And generally, as first pointed out by Mr Swinden, with 

 2c men, c of them, occupying initially the consecutive places 

 numbered c, c + 1, ..., 2c -1, will be picked out, if h is the 

 L. c. M. of c 4- 1, c + 2, . . . , 2c — 1 ; and the other c will be picked 

 out if k = h+l, though it may well be that there is a simpler 

 solution for another initial arrangement. It may be impossible 

 to arrange the men so that n specified individuals shall be 

 picked out in a defined order. 



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 + 98765 = 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 aa follows : 



The vessels can contain 24oz. 13 oz. lloz. 5oz. 



Their contents originally are... 24 . 



First, make their contents . 



Second, make their contents... 16 . 

 Third, make their contents ... 16. 

 Fourth, make their contents. . . 3 . 



Fifth, make their conten ts 3 . 



Lastly, make their contents ... 8 . 



