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
To begin, A must be a factor of 210 = 2*3*5*7
To see the structure of the triples suppose A=10
Then B=10n and C=10m so
10+10n+10m=210
1+n+m=21
n+m=20
for which there are 19 possibilities. This is just two less than 210/10 = 21
So for each factor
x of 210 there are two less triples than 210/
x. But 210/
x is just another factor of 210.
Procedure to count the number of triplets:
(sum of all factors of 210 except 1)2*(number of factors)
factors of 210 except 1: 2,3,5,7,6,10,14,15,21,35,105,70,42,30,210
They sum to 575 and there are 15 of them
5752*15=
545
This method will work for any value for the 3rd bullet, not just 210.

Posted by Jer
on 20140725 10:28:45 