2007 AMC 12B Problems/Problem 24

Problem 24

How many pairs of positive integers $(a,b)$ are there such that $gcd(a,b)=1$ and $\frac{a}{b}+\frac{14b}{9a}$ is an integer?

Combining the fraction, $\frac{9a^2 + 14b^2}{9ab}$ must be an integer.

Rewriting $b$ as $b = 3n$ for some positive integer $n$ , we can rewrite the fraction as $\frac{a^2 + 14n^2}{3an}$

Since the denominator now contains a factor of $n$ , we get $n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2$ .

For $b=3$ the original fraction simplifies to $\frac{a^2 + 14}{3a}$ .

For that to be an integer, $a$ must divide $14$ , and therefore we must have $a\in\{1,2,7,14\}$ . Each of these values does indeed yield an integer.

Thus there are four solutions: $(1,3)$ , $(2,3)$ , $(7,3)$ , $(14,3)$ and the answer is $\mathrm {(A)}$