210 kirkman's school-girls problem [ch. X. 



arranged by any of the methods used for solving the case of 

 6<? + 3 girls, while of the latter a typical triplet is used on each 

 of those days, from which the remaining triplets on that day 

 are obtained by cyclical permutations. 



This method covers all cases except when n= 15, 21, 39; 

 and solutions by it for all values of n less than 200 have been 

 written out. Sets of all the triplets required can be definitely 

 determined. One way of doing this is by finding the primitive 

 roots of the prime factors of 4m + 2q + 1, though in the simpler 

 cases the triplets can be written down empirically without 

 much trouble. An advantage of this method is that solutions 

 of several cases are obtained by the same work. Suppose that 

 we have arranged suitable triangles in a circle, having on 

 its circumference 12m + p or 3c equidistant points, and let y 

 be the greatest integer satisfying the indeterminate equation 

 2x + iy + 1 = c, where x = or x = 1, and a the highest multiple 

 of 6 included in x + y, then solutions of not less than y + 1 — a 

 cases can be deduced. Thus from a 27 circle arrangement 

 where c = 9, y = 2, x = 0, a = 0, we can by this method deduce 

 three solutions, namely when n = 57, 69, 81 ; from a 39 circle 

 arrangement where c = 13, y = 3, « = 0, ot = 0, we can deduce 

 four solutions, namely when n=81, 93, 105, 117. 



I have no space to describe the method fully, but I will give 

 solutions for two cases, namely for 33 girls (n = 33, m = 1, p = 3) 

 where there are 3 foci, and for 51 girls (n. = 51, m = 1, p = 9) 

 where there are 9 foci. 



For 33 girls we have 3 foci which we may denote by a, b, c, 

 and 30 points which we may denote by the numbers 1 to 30 

 placed at equidistant intervals on the circumference of a circle. 

 Then if the arrangement on the first day is (a. 5. 10), (b. 20. 25), 

 (c. 15. 30), (1. 2. 14), (16. 17. 29), (4 23. 26), (19. 8. 11), (9. 7. 3), 

 (24. 22. 18), (6. 27. 13), (21. 12. 28), a two-step cyclical per- 

 mutation of the numbers gives arrangements for 15 days; 

 that on the second day being (a. 7. 12), (6. 22. 27), &c. The 

 arrangement on the 16th day is (a. b. c), (1. 11. 21), (2. 12. 22), 

 (3. 13. 23), ...(10. 20.30). 



For 51 girls we have 9 foci which we may denote by a, b, c, 



