Find the number of triplets (A, B, C) of positive integers such that:
 A divides B, and:
 A divides C, and:
 A+B+C=210
As A divides both B and C it must divide A+B+C=210. Then A, B and C are all multiples of A with A being the smallest. B can be anything from 1*A through 2102*A at intervals of A, as neither B nor C=210AB may be zero. This amounts to 210/A  2 ways for each possible A:
A ways
1 208
2 103
3 68
5 40
6 33
7 28
10 19
14 13
15 12
21 8
30 5
35 4
42 3
70 1

545 total ways
Edit: "(2102*A)/A" corrected to "2102*A at intervals of A".
Edited on July 26, 2014, 9:38 am

Posted by Charlie
on 20140725 12:41:16 