AIME II Problem 2 2007

[zefuri]

Find the number of ordered triple (a,b,c) where a, b, and c are positive integers, a is a factor of b, a is a factor of c, and a+b+c=100.

[MistaPotta]

a is a factor of b means b = ma, for integer k.
a is a factor of c means c = na for integer k.

So a + b + c = a + ma + na = a (1 + m + n) = 100
Thus a and (1 + m + n) must be factors of 100, i.e., 1, 2, 4, 5, 10, 20, 25, 50, 100.

If a = 100, then 1 + m + n = 1, which isn't possible.

If a = 50 then 1 + m + n = 2, which isn't possible.

If a = 25 then 1 + m + n = 4, which has two possibilities (m,n) = (1,2) or (2,1)

If a = 20 then 1 + m + n = 5 - three possibilities
If a = 10 then 1 + m + n = 10 - eight possibilities
If a = 5 then 1 + m + n = 20 - eighteen possibilities
If a = 4 then 1 + m + n = 25 - twenty-three possibilities
If a = 2 then 1 + m + n = 50 - forty-eight possibilities
If a =1 then 1 + m + n = 100 - ninety-eight possibilities.

[stupidityismygam]