Difference between revisions of "2007 AMC 12B Problems/Problem 24"

Revision as of 09:52, 4 July 2013

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 $9$ , $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$

Rewriting $b$ as $b = 3n$ for some positive integer $n$ , we can rewrite the fraction 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 $\mathrm {(A)}$

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

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 ==See Also==
 {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}
+{{MAA Notice}}