Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2007 AMC 10B Problems/Problem 25

Revision as of 16:47, 18 September 2020 by Brian22 (talk | contribs) (→‎Problem: fixed some LaTeX)

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

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 $\frac{7}{3}$.

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_25&oldid=133810"