198 kiekman's school-girls problem [ch. X 



of triangles satisfies the conditions: (4.11.25), (5.8.23), 

 (6. 7. 16), (9. 13. 15). The complementary system is (17. 24. 12), 

 (18. 21. 10), (19. 20, 3), (22. 26. 2). These triplets with 

 (&. 1. 14) give an arrangement for the first day; and, by 

 rotating the system cyclically, the arrangements for the re- 

 maining 12 days can be found immediately. 



I proceed to give one solution of this type for every remaining case where n 

 is less than 100. From the result the triads or groupings used can be obtained. 

 It is sufficient in each case to give an arrangement on the first day, since the 

 arrangements on the following days are at once obtainable by cyclical per- 

 mutations. 



I take first the three cases, 33, 57, 81, where n is of the form 24m + 9. 

 In these cases the arrangements on the other days are obtained by one-step 

 cyclical permutations. 



For 33 girls, a solution is given by the system of triplets (2. 11. 16), 

 (4. 6. 10), (5. 13. 30), (7. 8. 19), (9. 28. 31), and the complementary system 

 (18. 27. 32), (20. 22. 26), (21. 29. 14), (23. 24. 3), (25. 12. 15). These 10 triplets, 

 together with that represented by (ft. 1. 17), will give an arrangement for the 

 first day. 



For 57 girls, a possible arrangement of triplets is (18. 13. 50), (20. 11. 28), 

 (21. 52. 3), (8. 10. 51), (4. 25. 26), (2. 6. 12), (7. 19. 33), (27. 43. 16), (37, 14. 17). 

 These, with the 9 complementary triplets, and the diameter triplet (1. ft. 29), 

 give an arrangement for the first day. 



For 81 girls an arrangement for the first day consists of the diameter triplet 

 (1. ft. 41), the 13 triplets (3. 35. 42), (4. 10. 29), (5. 28. 56), (6. 26. 39), (7. 15. 17), 

 (8. 11. 32), (13. 27 49), (14. 19. 30), (20. 37. 38), (21. 25. 52), (24. 36. 62), 

 (18. 33. 63), (31. 40. 74), and the 13 complementary triplets. 



I take next the three cases, 51, 75, 99, where n is of the form 24m+3. In 

 these cases the arrangements on the other days are obtained either by one-step 

 or by two-step cyclical permutations. 



For 51 girls, an arrangement for the first day consists of the diameter 

 triplet (ft. 1. 26), the 8 triplets (2. 9. 36), (4. 7. 25), (6. 10. 19), (8. 14. 22), 

 (12. 17. 45), (13. 24. 48), (15. 16. 46), (18. 28. 30), and the 8 complementary 

 triplets. 



For 75 girls, an arrangement for the first day consists of the diameter triplet 

 (ft. 1. 38), the 12 triplets (2. 44. 55), (4. 11. 19), (5. 50. 66), (6. 52. 57), 

 (8. 46. 58), (10. 59. 65), (12. 60. 64), (14. 24. 68), (16. 25. 72), (17. 3. 74), 

 (33. 34. 73), (30. 32. 63), and the 12 complementary triplets. 



For 99 girls an arrangement for the first day consists of the diameter 

 triplet (1. ft. 50), the 16 triplets (2. 17. 47), (3. 9. 68), (4. 44. 82), (5. 12. 75), 

 (6. 32. 42), (7. 23. 97), (8. 21. 30), (15. 20. 76), (16. 35. 85), (18. 45. 62), 

 (22. 40. 63), (25. 37. 92), (28. 29. 80), (34. 38. 69), (39. 41. 73), (46. 49. 60), 

 and the 16 complementary triplets. 



It is also possible to obtain, for numbers of the form 24m + 3, solutions 

 which are uniquely two-step, but in these the complementary triangles are not 



