2007 AIME II Problems/Problem 2

Problem

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

Solution

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

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

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

2007 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2007 AIME II Problems/Problem 2

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