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== Problem ==
Find the number of ordered triples <math>(a,b,c)</math> where <math>a</math>, <math>b</math>, and <math>c</math> are positive [[integer]]s, <math>a</math> is a [[factor]] of <math>b</math>, <math>a</math> is a factor of <math>c</math>, and <math>a+b+c=100</math>.

== Solution ==
Denote <math>x = \frac{b}{a}</math> and <math>y = \frac{c}{a}</math>. The last condition reduces to <math>\displaystyle a(1 + x + y) = 100</math>. Therefore, <math>\displaystyle 1 + x + y</math> is equal to one of the 9 factors of <math>\displaystyle 100 = 2^25^2</math>.

Subtracting the one, we see that <math>\displaystyle x + y = \{0,1,3,4,9,19,24,49,99\}</math>. There are exactly <math>\displaystyle n - 1</math> ways to find pairs of <math>\displaystyle (x,y)</math> if <math>\displaystyle x + y = n</math>. Thus, there are <math>\displaystyle 0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = 200</math> solutions of <math>(a,b,c)</math>.

Alternatively, note that the sum of the divisors of <math>100</math> is <math>(1 + 2 + 2^2)(1 + 5 + 5^2) \displaystyle</math> (notice that after distributing, every divisor is accounted for). This evaluates to <math>7 \cdot 31 = 217</math>. Subtract <math>9 \cdot 2</math> for reasons noted above to get <math>199</math>. Finally, this changes <math>\displaystyle 1 \Rightarrow -1</math>, so we have to add one to account for that. We get <math>200</math>.

== See also ==
{{AIME box|year=2007|n=II|num-b=1|num-a=3}}

[[Category:Intermediate Algebra Problems]]