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

Revision as of 15:13, 5 January 2009

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

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

@@ Line 8: / Line 8: @@
 Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer.
-Since the denominator contains a factor of <math>9</math>,
+Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math>
-<math>9 | 9a^2 + 14b^2</math>
+Rewriting <math>b</math> as <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction as <math>\frac{a^2 + 14n^2}{3an}</math>
-<math>9 | b^2</math>
+Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>.
-<math>3 | b</math>
+But since <math>1=gcd(a,b)=gcd(a,3n)=gcd(a,n)</math>, we must have <math>n=1</math>, and thus <math>b=3</math>.
-Rewriting <math>b</math> as <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction <math>\frac{a^2 + 14n^2}{3an}</math>
+For <math>b=3</math> the original fraction simplifies to <math>\frac{a^2 + 14}{3a}</math>.
-Since the denominator now contains a factor of <math>n</math>,
+For that to be an integer, <math>a</math> must divide <math>14</math>, and therefore we must have <math>a\in\{1,2,7,14\}</math>. Each of these values does indeed yield an integer.
-<math>n | a^2 + 14n^2</math>
+Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm {(A)}</math>
-<math>n | a^2</math>
-<math>n | a</math>
-Rewriting <math>a</math> as <math>a = mn</math> for some positive integer <math>m</math>, we can rewrite the fraction again as <math>\frac{m^2 + 14}{3m}</math>
-Since the denominator contains <math>m</math>,
-<math>m | m^2 + 14</math>
-<math>m | 14</math>
-<math>m \in (1,2,7,14)</math>
-Checking back to the fraction, all of these <math>m</math> do indeed yield integers.
-Now, returning to <math>a</math> and <math>b</math>
-<math>a \in (n,2n,7n,14n)</math> and
-<math>b = 3n</math>
-Since <math>gcd(a,b) = 1</math>, <math>n</math> must be <math>1</math>.
-This yields four possible pairs
-<math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math>
-<math>4 \Rightarrow \mathrm {(A)}</math>
 ==See Also==
 {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}

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

Revision as of 15:13, 5 January 2009

Problem 24

Solution

See Also