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 210-2*A at intervals of A, as neither B nor C=210-A-B 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
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545 total ways
Edit: "(210-2*A)/A" corrected to "210-2*A at intervals of A".
Edited on July 26, 2014, 9:38 am
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Posted by Charlie
on 2014-07-25 12:41:16 |
solution
