Problem:
Let N be the number of ordered triples (A,B,C) of integers satisfying the conditions
(a) 0≤A<B<C≤99,
(b) there exist integers a,b, and c, and prime p where 0≤b<a<c<p,
(c) p divides A−a,B−b, and C−c, and
(d) each ordered triple (A,B,C) and each ordered triple (b,a,c) form arithmetic sequences.
Find N.
Solution:
Let d=a−b=c−a. Then there are integers m and n so that np−d= B−A=C−B=mp+2d, implying that 3d=(n−m)p. Note that 2d<p, so 3 cannot divide n−m. It follows that p=3, and (b,a,c)=(0,1,2). Therefore the ordered triples that meet the conditions are precisely those of the form (1+3j,3+3j+3k,5+3j+6k), where j≥0,k≥0, and j+2k≤31. Thus
N=k=0∑15(32−2k)=16⋅32−2(215⋅16)=272
The problems on this page are the property of the MAA's American Mathematics Competitions