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



may be required. The process depends on finding primitive roots 

 of the prime factors of whatever number is taken as the base 

 of the solution. When (w — 1)/2 is prime, of the form Qu + 1, 

 and is taken as base, the order is obtainable at once, and the 

 rules for grouping the numbers are easy of application ; owing 

 to considerations of space I here confine myself to such instances, 

 but similar though somewhat longer methods are applicable 

 to all cases. 



I use the geometrical representation already explained. 

 We have n = 1y + 1, and y is a prime of the form 6m + 1. In 

 forming the triplets we either proceed directly by arranging all 

 the points in threes, or we arrange some of them in pairs and 

 make the selection of the third point dependent on those of the 

 two first chosen, leaving only a few triplets to be obtained 

 otherwise. In the former case we have to arrange the numbers 

 in triplets so that each difference will appear twice, and so that 

 no two differences will appear together more than once. In 

 the latter case we have to arrange the numbers so that the 

 differences between the numbers in each pair comprise con- 

 secutive integers from 1 upwards and are all different. In both 

 cases, we commence by finding a primitive root of y, say x. The 

 residues to the modulus y of the 6m successive powers of # form 

 a series of numbers, el, e2, e3, . . . , comprising all the integers from 

 1 to 6m, and when taken in the order of the successive powers, 

 they can be arranged in the manner required by definite rules. 



I will apply the method to the case of 27 girls from which 

 the general theory, in the restricted case where y is a prime 

 of the form 6m + 1, will be sufficiently clear. In this case 

 we have w=27, y=13, w=2, and x = 2. I take 13 as the 

 base of the analysis. I will begin by pairing the points, and 

 this being so, it is convenient to represent the girls by a point 

 k at the centre of a circle and points ol, 61, a2, 62, ... at 

 equidistant intervals on the circumference. We reserve k, ol3, 

 613 for one triplet, and we have to arrange the other 24 points 

 so as to form 8 triangles of certain types. The residues are in 

 the order 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1, and these may be 

 taken as the suffixes of the remaining ' a 's and ' 6 's. 



