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 Testy Triplet Treat II (Posted on 2014-07-25)
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 No Rating

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 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   2810   1914   1315   1221    830    535    442    370    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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