CH. x] kirkman's school-girls problem 207 



(23. 31. 69), (54. 62. 7), (85. 93. 38), (22. 53. 84). From each of the 15 triplets 

 (1. 14. 42), (2. 15. 43), (3. 16. 44), (1. 33. 44), (2. 34. 45), (3. 35. 46), (1. 11. 30), 

 (2. 12. 31), (3. 13. 32), (1. 6. 41), (2. 7. 42), (3. 8. 43), (1. 15. 35), (2. 16. 36), 

 (3. 17. 37), an arrangement for one of the remaining 15 days is obtainable by a 

 three-step oyclical permutation. 



Before leaving the subject of numbers of this type I give two other solutions 

 of the case when n = 21, one to illustrate the use of the suffix notation, and the 

 other a cyclical solution which is uniquely three-step. 



If we employ the suffix notation, the suffixes, with the type here used, are 



somewhat trying to read. Accordingly hereafter I shall write al, a2 instead 



of a„ a a In the case of 21 girls, an arrangement for the first day is 



(al. a2. a4), (61. 62. 64), (el. c2. ci), (a3. 66. c5), (63. c6. a5), (c3. a6. 65), 

 (a7. 67. c7). From this, by one-step oyclical permutations of the suffixes, we 

 get arrangements for the 2nd, 3rd, 4th, 5th, 6th and 7th days. The arrange- 

 ment for the 8th day can be obtained from the triplet (al. 62. c4) by permuting 

 the suffixes cyclically one step at a time. Similarly the arrangement for the 

 9th day can be obtained from the triplet (61. c2. a4) and that for the 10th day 

 from the triplet (<sl. a2. 54). Thus with seven suffixes we keep 7 for each symbol 

 in one triplet, and every other triplet depends on one or other of only two 

 arrangements, namely, (1. 2. 4), or (3. 6. 5). If the solution be written out 

 at length the principle of the method used will be clear. 



Cyclical solutions which are uniquely three-step can also be obtained for 

 numbers of the form 18m +3; in them the same triads can be used as 

 before, but they are not placed at equal intervals in the circle. I give 21 girls 

 as instance. The arrangements on the first 7 days can be obtained from the 

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

 (5. 12. 19) by a three-step (but not by a one-step) cyclical permutation. From 

 each of the triplets (1. 3. 11), (2. 4. 12), (3. 5. 13), an arrangement for one of 

 the remaining 3 days is obtainable. 



Next, I consider those cases where n is of the form 18m + 9. Here, regular 

 solutions in the suffix notation can be obtained in all cases except in that 

 of 27 girls, but if the same solutions are expressed in the numerical notation, 

 the triads are irregular. Accordingly, except when n = 27, it is better to use the 

 suffix notation. I will deal with the case when n=27 after considering the 

 cases when n =45, 63, 81, 99. 



For 45 girls, an arrangement for tho first day consists of the 5 triplets 

 (al. al2. al3), (a2. a9. all), (a5. alO. 615), (a4. 63. c7), (a8. 66. cl4), and the 

 10 analogous triplets, namely, (61. 412. 613), (el. cl2. cl3), (62. 69. 611), 

 (c2. c9. ell), (65. 610. cl5), (c5. clO. al5), (64. c3. a7), (ci. a3. 67), (68. c6. al4), 

 (c8. a6. 614). From these, by one-step cyclical permutations of the suffixes, the 

 arrangements for 15 days can be got. Each of the 2 triplets (ci. 63. a7), 

 (c8. 66.al4), the 4 analogous triplets, namely, (a4.e3.67), (64. a3.c7), (a8.c6. 614), 

 (58. a6. cl4), and the triplet (al. 61. cl), gives, by a one-step cyclical permuta- 

 tion of the suffixes, an arrangement for one of the remaining 7 days. 



For 63 girls, an arrangement for the first day consists of the 7 triplets 

 (al. alO. a9), (a5. aS. a3), (a4. al9. al5), (a7. al4. 621), (a20. 611. cl2), 

 («16. 613. cl8), (al7. 52. c6), and the 14 analogous triplets, From these, by 



