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



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

 is less than 100. I give some in the numerical, others in the suffix notation. 

 The results will supply an indication of the process used. 



First, I consider those cases where n is of the form 18m + 3. In these cases 

 it is always possible to find 2m triads each repeated thrice, and one equilateral 

 triad, and to use the equilateral and m triads in the first part of the solution, 

 the 3 triplets representing any one of these m triads being placed in the circle 

 at equal intervals from each other in the first day's arrangement. From this, 

 three-step or one-step cyclical permutations give arrangements for 6m + 1 days 

 in all. In the second part of the solution each of the m remaining triads 

 is used thrice ; it suffices to give the first triplet on each day, since from it the 

 other triplets on that day are obtained by a three-step cyclical permutation. 



For 21 girls an arrangement for the first day, for the first part of the solution, 

 is (1. 4. 10), (8. 11. 17), (15. 18. 3), (2. 6. 7), (9. 13. 14), (16. 20. 21), (5. 12. 19). 

 From this, three-step or one-step cyclical permutations give arrangements for 

 7 days in all. The first triplets used in the second part of the solution are 

 (1. 3. 11), (2. 4. 12), (3. 6. 13). Each of these three triplets gives by a three- 

 step cyclical permutation an arrangement for one of the remaining 3 days. 



For 39 girls, arrangements for 13 days can be obtained from the following 

 arrangement of triplets for the first day: (1. 4. 13), (14. 17. 26), (27. 30. 39), 

 (2. 8. 23), (15. 21. 36), (28. 34. 10), (7. 11. 12), (20. 24. 25), (33. 37. 38), (3. 5. 19), 

 (16. 18. 32), (29. 81. 6), (9. 22. 35). From eaoh of the 6 triplets (1. 8. 18), 

 (2. 9. 19), (3. 10. 20), (1. 9. 20), (2. 10. 21), (3. 11. 22), an arrangement for one 

 of the remaining 6 days is obtainable. 



For 57 girls, arrangements for 19 days can be obtained from the following 

 arrangement of triplets for the first day: (1. 4. 25), (20. 23. 44), (39. 42. 6), 

 (2. 8. 17), (21. 27. 36), (40. 46. 55), (3. 15. 33), (22. 34. 52), (41. 53. 14), 

 (18. 19. 26), (37. 38. 45), (56. 67. 7), (13. 30. 35), (32. 49. 54), (51. 11. 16), 

 (9. 29. 43), (28. 48. 5), (47. 10. 24), (12. 31. 50). From each of the 9 triplets 

 (1. 3. 14), (2. 4. 15), (3. 5. 16), (1. 5. 30), (2. 6. 31), (3. 7. 32), (1. 11. 27), 

 (2. 12. 28), (3. 13. 29), an arrangement for one of the remaining 9 days is 

 obtainable. 



For 75 girls, arrangements for 25 days can be obtained from the following 

 arrangement of triplets for the first day: (1. 4. 10), (26. 29. 35), (51. 54. 60), 

 (3. 15. 36), (28. 40. 61), (53. 65. 11), (2. 17. 41), (27. 42. 66), (52. 67. 16), 

 (13. 31. 58), (38. 56. 8), (63. 6. 33), (14. 21. 22), (39. 46. 47), (64. 71. 72), 

 (7. 18. 20), (32. 43. 45), (57. 68. 70), (5. 9. 37), (30. 34. 62), (55. 59. 12), 

 (19. 24. 50), (44. 49. 75), (69. 74. 25), (23. 48. 73). From eaoh of the 12 triplets 

 (1. 11. 30), (2. 12. 31), (3. 13. 32), (1. 15. 35), (2. 16. 36), (3. 17. 37), (1. 17. 39), 

 (2. 18. 40), (3. 19. 41), (1. 18. 41), (2. 19. 42), (3. 20. 43), an arrangement for 

 one of the remaining 12 days is obtainable. 



For 93 girls, arrangements for 31 days can be obtained from the following 

 arrangement of triplets for the first day : (1. 76. 79), (32. 14. 17), (63. 45. 48), 

 (13. 25. 55), (44. 56. 86), (75. 87. 24), (29. 50. 74), (60. 81. 12), (91. 19. 43), 

 (20. 26. 59), (51. 57. 90), (82. 88. 28), (3. 30. 39), (34. 61. 70), (65. 92. 8), 

 (27. 64. 71), (58. 2. 9), (89. 33. 40), (35. 36. 52), (66. 67. 83), (4. 5. 21), 

 (11. 15. 37), (42. 46. 68), (73. 77. 6), (16. 18. 41), (47. 49. 72), (78. 80. 10), 



