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