# Difference between revisions of "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? $\mathrm {(A)} 4$ $\mathrm {(B)} 6$ $\mathrm {(C)} 9$ $\mathrm {(D)} 12$ $\mathrm {(E)} \text{infinitely many}$

## Solution

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

Since the denominator contains a factor of $9$, $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$

Since $b = 3n$ for some positive integer $n$, we can rewrite the fraction(divide by $9$ on both top and bottom) as $\frac{a^2 + 14n^2}{3an}$

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

But since $1=gcd(a,b)=gcd(a,3n)=gcd(a,n)$, we must have $n=1$, and thus $b=3$.

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 $\mathrm {(A)}$

## Solution 2

Let's assume that $\frac{a}{b} + \frac{14b}{9a} = m$ We get-- $9a^2 - 9mab + 14b^2 = 0$

Factoring this, we get $4$ equations- $(3a-2b)(3a-7b) = 0$ $(3a-b)(3a-14b) = 0$ $(a-2b)(9a-7b) = 0$ $(a-b)(9a-14b) = 0$

(It's all negative, because if we had positive signs, $a$ would be the opposite sign of $b$)

Now we look at these, and see that- $3a=2b$ $3a=b$ $3a=7b$ $3a=14b$ $a=2b$ $9a=7b$ $a=b$ $9a=14b$

This gives us $8$ solutions, but we note that the middle term needs to give you back $9m$.

For example, in the case $(a-2b)(9a-7b)$, the middle term is $-25ab$, which is not equal by $-9m$ for whatever integar $m$.

Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total $\mathrm {(A)}$

## Solution 3

Let $u = \frac{a}{b}$. Then the given equation becomes $u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}$.

Let's set this equal to some value, $k \Rightarrow \frac{9u^2 + 14}{9u} = k$.

Clearing the denominator and simplifying, we get a quadratic in terms of $u$: $9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}$

Since $a$ and $b$ are integers, $u$ is a rational number. This means that $\sqrt{(9k)^2 - 504}$ is an integer.

Let $\sqrt{(9k)^2 - 504} = x$. Squaring and rearranging yields: $(9k)^2 - x^2 = 504$ $(9k+x)(9k-x) = 504$.

In order for both $x$ and $a$ to be an integer, $9k + x$ and $9k - x$ must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let $9k + x = 2m$ and $9k - x = 2n$.

Then: $2m \cdot 2n = 504$ $mn = 126$.

Factoring 126, we get $6$ pairs of numbers: $(1,126), (2,63), (3,42), (6,21), (7,18) & (9,14)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.).

Looking back at our equations for $m$ and $n$, we can solve for $k = \frac{2m + 2n}{18} = \frac{m+n}{9}$. Since $k$ is an integer, there are only $2$ pairs of $(m,n)$ that work: $(3,42) & (6,21)$ (Error compiling LaTeX. ! Misplaced alignment tab character &.). This means that there are $2$ values of $k$ such that $u$ is an integer. But looking back at $u$ in terms of $k$, we have $\pm$, meaning that there are $2$ values of $u$ for every $k$. Thus, the answer is $2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}$.

## See Also

 2007 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS