CH. x] kirkman's school-girls problem 215 



of the 'a's on the 13 circle. This gives the following solution, 

 in which the first six triangles are isosceles: (a 2. a 8. 65), 

 (a4. a3. 610), (06. all. 62), (al2. a9. 64), (o5. at. 66), 

 (alO. al. 612), (63. 69. 61), (611. 67. 68), (*. al3. 613). 



In the case of 27 girls, we may equally well represent the 

 points hy k at the centre of the circle and 26 equidistant points 

 1, 2, ... 26 on the circumference. The points previously denoted 

 by a and 6 with the suffix h are now denoted by the numbers 

 2h— 1 and 2h. Hence the basic pairs (2. 8), (4. 3), ... become 

 (3. 15), (7. 5), (11. 21), (23. 17), (9. 13), (19. 1), and the corre- 

 sponding scalene arrangement for the first day is (3. 15. 20), 

 (7. 5. 14), ... (k. 25. 26). From this by a two-step cyclical 

 permutation of the numbers, an arrangement for 13 days can 

 be got. 



The case of 27 girls can also be treated by the direct 

 formation of triplets. The triplets must be such that each 

 difference is represented twice, but so that the groups of 

 differences are different. There are analytical rules for forming 

 such triplets somewhat analogous to those I have given for 

 forming basic pairs, but their exposition would be lengthy, 

 and I will not discuss them here. One set which will answer 

 our purpose is (1. 12. 5), (2. 3. 10), (4. 6. 9), (8. 11. 7), giving 

 respectively the differences [2, 6, 4], [1, 6, 5], [2, 3, 5], [3, 4, 1]. 

 Now every difference d in a 13 circle will correspond to d 

 or 13 — d in a 26 circle, and every residue e in a 13 circle 

 will correspond to e or 13 + e in a 26 circle. Further, a 

 triplet in the 26 circle must either have three even differences, 

 or one even and two odd differences. Hence from the above 

 sets we can get the following arrangement for the first day, 

 (1.25.5) and (14.12.18) with differences [2,6,4], (15.3.10) 

 and (2. 16. 23) with differences [12, 7, 5], (17. 6. 9) and (4. 19. 22) 

 with differences [11, 3, 8], (21. 11. 20) and (8. 24. 7) with 

 differences [10, 9, 1], and (k 13. 26). From this by either a 

 one-step or a two-step cyclical permutation of the numbers, an 

 arrangement for 13 days can be got. I will not go into further 

 details about the deduction of other similar solutions. A similar 

 method is always applicable when n is of the form 24m + 3. 



