216 



KIRKMANS SCHOOL-GIRLS PROBLEM 



[CH. X 



The process by pairing when y is a prime of the form 6w+l 

 is extremely rapid. For instance, in the case of 15 girls we 

 have n = 15, y = 7, u = 1, tc= 5. The order of the residues is 

 5, 4, 6, 2, 3, 1. By our rule we can at once arrange basic pairs 

 (5. 4), (6. 2), (3. 1). From these pairs we can obtain numerous 

 solutions. Thus using scalene triangles as above explained, we 

 get as an arrangement for the first day (a5, a4, 62), (a6. a 2. 61), 

 (a3. ol. 64), (63. 65. 66), (k. a7. 67), from which by a one-step 

 cyclical permutation of the numbers, arrangements for the seven 

 days can be obtained. Using the basic pairs as bases of 

 isosceles triangles, we get as an arrangement for the first day 

 (a5. a4. 61), (a6. a% 64), (a3. al. 62), (63. 65. 66), (k a?. 67). 



Again, take the case of 39 girls. Here we have n = 39, 

 y=19, m=3, #=3. The order of the residues is 3, 9, 8; 5, 15, 7; 



2, 6, 18 ; 16, 10, 11 ; 14, 4, 12 ; 17, 13, 1. The basic pairs are 

 (3. 5), (9. 15), (8. 7), (2. 16), &c. These are the suffixes of the 

 'a's. The possible triplets which determine what '6's are to 

 be associated with these, and what '6's are to be left 

 for the remaining three triangles, can be determined as 

 follows : From the residue series we obtain the derivative series 



3, 9, 8, 5, 4, 7, 2, 6, 1, &c. The divisions are (i) 3, 5, 2 ; (ii) 9, 4, 6 ; 

 (iii) 8, 7, 1. A cyclical arrangement like that given above 

 leads to the result in the left half of the annexed table which 

 does not satisfy our condition. A .cyclical displacement of the 

 symbols in the vertical lines in the first column leads to the 

 arrangement given in the right half of the table, and shows 



that (3. 15. 18), (5. 6. 11), &c. are possible triplets. From 

 these results numerous solutions can be deduced in the same 

 way as above. For instance one solution is (a 3. <x5. 64), 



