202 kirkman's school-girls problem [ch. x 



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

 n la less than 100. In each case I give an arrangement on the first day; the 

 arrangements for the other days oan be got from it by a two-step cyclical 

 permutation of the numbers. 



In the case of 27 girls, one arrangement on the first day is (ft. 1. 14), 

 (19. 21. 20), (3. 9. 6), (13. 23. 18), (5. 17. 24), (7. 25. 16), (11. 15. 26), (10. 4. 2), 

 (12. 22. 8). 



In the case of 39 girls, one arrangement for the first day is (ft. 1. 20), 

 (35. 37. 36), (7. 13. 10), (19. 29. 24), (11. 25. 18), (3. 21. 12), (17. 33. 6), 

 (15. 27. 2), (23. 31. 8), (5. 9. 26), (4. 16. 32), (14. 34. 38), (22. 28. 30). 



In the case of 51 girls, one arrangement for the first day is (ft. 1. 26), 

 (21. 23. 22), (11. 17. 14), (35. 45. 40), (25. 39. 32), (15. 33. 24), (13. 41. 2), 

 (5. 31. 18), (19. 49. 34), (27. 43. 10), (9. 47. 28), (29. 37. 8), (3. 7. 30), 

 (4. 20. 46), (6. 36. 42), (12. 16. 44), (38. 48. 50). 



In the case of 63 girls, one arrangement for the first day is (ft. 1. 32), 

 (57. 59. 58), (23. 29. 26), (37. 47. 42), (5. 53. 60), (17. 61. 8), (3. 43. 54), 

 (7. 33. 20), (9. 39. 24), (27. 55. 10), (21. 45. 2), (11. 31. 52), (35. 51. 12), 

 (13. 25. 50), (41. 49. 14), (15. 19. 48), (16. 18. 36), (6. 34. 38), (46. 56. 62), 

 (22. 30. 44), (4. 28. 40). 



In the case of 75 girls, one arrangement for the first day is (ft. 1. 38), 

 (23. 25. 24), (3. 9. 6), (29. 39. 34), (7. 21. 14), (51. 69. 60), (33. 55. 44), 

 (11. 59. 72), (35. 65. 50), (37. 71. 54), (27. 63. 8), (15. 57. 36), (13. 41. 64), 

 (43. 67. 18), (19. 73. 46), (31. 47. 2), (5. 17. 48), (53. 61. 20), (45. 49. 10), 

 (30. 32. 52), (22. 26. 66), (56. 62. 70), (12. 28. 74), (16. 40. 58), (4. 42. 68). 



In the case of 87 girls, one arrangement for the first day is (ft. 1. 44), 

 (61. 63. 62), (73. 79. 76), (35. 45. 40), (11. 83. 4), (25. 43. 34), (59. 81. 70), 

 (7. 33. 20), (41. 71. 56), (23. 75. 6), (27. 65. 46), (13. 57. 78), (5. 51. 28), 

 (3. 39. 64), (15. 47. 74), (9. 67. 38), (31. 55. 86), (19. 85. 52), (21. 37. 72), 

 (17. 29. 66), (69. 77. 30), (49. 53. 8), (10. 12. 24), (22. 26. 84), (18. 54. 60), 

 (2. 50. 80), (32. 42. 58), (14. 36. 82), (16. 48. 68). 



In the case of 99 girls, one arrangement for the first day is (ft. 1. 50), 

 (47. 49. 48), (53. 59. 56), (55. 65. 60), (57. 71. 64), (23. 41. 32), (17. 39. 28), 

 (63. 89. 76), (5. 35. 20), (3. 67. 84), (9. 69. 88), (29. 85. 8), (27. 79. 4), 

 (25. 73. 98), (33. 77. 6), (21. 61. 90), (45. 81. 14), (51. 83. 18), (15. 43. 78), 

 (13. 37. 74), (11. 31. 70), (75. 91. 34), (7. 19. 62), (87. 95. 42), (93. 97. 46), 

 (58. 94. 96), (40. 44. 66), (10. 16. 26), (30. 38. 82), (68. 80. 2), (22. 36. 92), 

 (24. 54. 72), (12. 52. 86). 



This method may be also represented as a one-step cycle. 

 For if we denote the girls by a point k at the centre of the 

 circle, and points a lt b l3 a. 2 , b.^, a 3 , b 3 , ... placed in that order 

 on the circumference, we can re-write the solutions in the 

 suffix notation, and then the cyclical permutation of the 

 numbers denoting the suffixes is by one step at a time. 



The one-step and two-step methods described above cover 



