2007 AMC 10B Problems/Problem 25

How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:

$\frac{a}{b} + \frac{14b}{9a}$

is an integer?

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

Solution

Let $x = \frac{a}{b}$ . We can then write the given expression as $x+\frac{14}{9x} = k$ where $k$ is an integer. We can rewrite this as a quadratic, $9x^2 - 9kx + 14 = 0$ . By the Quadratic Formula, $x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}$ . We know that $x$ must be rational, so $9k^2-56$ must be a perfect square. Let $9k^2-56 = n^2$ . Then, $56 = 9k^2-n^2 = (3k - n)(3k + n)$ . The factors pairs of $56$ are $1$ and $56$ , $2$ and $28$ , $4$ and $14$ , and $7$ and $8$ . Only $2$ and $28$ and $4$ and $14$ give integer solutions, $k = 5$ and $n = 13$ and $k = 3$ and $n = 5$ , respectively. Plugging these back into the original equation, we get $\boxed{4 (A)}$ possibilities for $x$ , namely $\frac{1}{3}, \frac{14}{3}, \frac{2}{3},$ and $\frac{7}{3}$ .

2007 AMC 10B (Problems • Answer Key • Resources)
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2007 AMC 10B Problems/Problem 25

Solution

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