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?

$\mathrm {(A)} 4$ $\mathrm {(B)} 6$ $\mathrm {(C)} 9$ $\mathrm {(D)} 12$ $\mathrm {(E)} \text{infinitely many}$

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

Since the denominator contains a factor of $9$ ,

$9 | 9a^2 + 14b^2$

$9 | b^2$

$3 | b$

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

Since the denominator now contains a factor of $n$ ,

$n | a^2 + 14n^2$

$n | a^2$

$n | a$

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

Since the denominator contains $m$ ,

$m | m^2 + 14$

$m | 14$

$m \in (1,2,7,14)$

Checking back to the fraction, all of these $m$ do indeed yield integers.

Now, returning to $a$ and $b$

$a \in (n,2n,7n,14n)$ and $b = 3n$

Since $gcd(a,b) = 1$ , $n$ must be $1$ . This yields four possible pairs $(1,3)$ , $(2,3)$ , $(7,3)$ , $(14,3)$

$4 \Rightarrow \mathrm {(A)}$

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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