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Testy Triplet Treat II (Posted on 2014-07-25) Difficulty: 3 of 5
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

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 3 of 4 |
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
    ---
    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
  Posted by Charlie on 2014-07-25 12:41:16

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