Problem:
Find the number of ordered triples (a,b,c) where a,b, and c are positive integers, a is a factor of b,a is a factor of c, and a+b+c=100.
Solution:
For any such ordered triple (a,b,c), because a is a factor of b+c,a is also a factor of 100. Thus a is an element of {1,2,4,5,10,20,25}, and ab and ac are positive integers for which ab+ac=a100−1 (Note that if a=50 or 100, then at least one of b and c is zero). Because ab and ac are positive integers, there are (for each choice of a ) a100−2 pairs ab and ac. Thus there are 1100+2100+4100+5100+10100+20100+25100−2⋅7=214−14=200 such triples.
The problems on this page are the property of the MAA's American Mathematics Competitions