Difference between revisions of "2007 AMC 10B Problems/Problem 25"

Revision as of 00:39, 25 December 2020

Problem

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

Also can refer to the | 2007 AMC 12B #24 (same problem)

Solution 1

We will use the divisibility notation ( $a|b$ ), which means $b$ is divisible by $a$ . Getting common denominators, we have to find coprime $(a,b)$ such that $9ab|(9a^2+14b^2)$ . $b$ is divisible by 3 because 14 is not a multiple of three in the equation, so $b$ must balance it and make them integers. Since $a$ and $b$ are coprime, $a|9a^2+14b^2 \implies a|14$ . Similarly, $b|9$ . However, $b$ cannot be $9$ as $81a|81 \cdot 14 + 9a^2$ only has solutions when $3|a$ . Therefore, $b=3$ and $a \in \{1,2,7,14\}$ . Checking them all (or noting that $4$ is the smallest answer choice), we see that they work and the answer is $\boxed{\mathrm{(A) \ } 4}$ .

Solution 2

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 $\boxed{\mathrm{(A) \ } 4}$ possibilities for $x$ , namely $\frac{1}{3}, \frac{14}{3}, \frac{2}{3},$ and $\frac{7}{3}$ .

@@ Line 11: / Line 11: @@
 ==Solution==
-Also can refer to the [https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24|2007 AMC 12B #24] (same problem)
+Also can refer to the [https://artofproblemsolving.com/wiki/index.php/2007_AMC_12B_Problems/Problem_24 | 2007 AMC 12B #24] (same problem)
 === Solution 1 ===

2007 AMC 10B (Problems • Answer Key • Resources)
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