2007 AIME II Problem 2

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2.

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

Answer: 200
Concepts:divisibilityDiophantine Equationcasework
Difficulty rating: 2070
Solution:

Since aa divides bb and c,c, it divides a+b+c=100.a + b + c = 100. Write b=asb = as and c=atc = at with s,t1;s, t \ge 1; then a(1+s+t)=100,a(1 + s + t) = 100, so s+t=100a1.s + t = \frac{100}{a} - 1. For positive ss and tt we need 100a3,\frac{100}{a} \ge 3, so a{1,2,4,5,10,20,25}.a \in \{1, 2, 4, 5, 10, 20, 25\}.

For each such a,a, the equation s+t=100a1s + t = \frac{100}{a} - 1 has 100a2\frac{100}{a} - 2 ordered positive solutions. Summing, (100+50+25+20+10+5+4)27=21414=200. \begin{aligned} &\small (100 + 50 + 25 + 20 + 10 + 5 + 4) \\ &\quad {}- 2 \cdot 7 \\ &= 214 - 14 = 200. \end{aligned}

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