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

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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 $\frac{a}{b} + \frac{14b}{9a} = m}$ (Error compiling LaTeX. Unknown error_msg) 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. Unknown error_msg).

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

@@ Line 69: / Line 69: @@
 Let <math>u = \frac{a}{b}</math>. Then the given equation becomes <math>u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}</math>.
-Let's set this equal to some value, <math>a \Rightarrow \frac{9u^2 + 14}{9u} = a</math>.
+Let's set this equal to some value, <math>k \Rightarrow \frac{9u^2 + 14}{9u} = k</math>.
 Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>:
-<math>9u^2 - 9au + 14 = 0 \Rightarrow u = \frac{9a \pm \sqrt{(9a)^2 - 504}}{18}</math>
+<math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math>
-Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9a)^2 - 504}</math> is an integer.
+Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9k)^2 - 504}</math> is an integer.
-Let <math>\sqrt{(9a)^2 - 504} = x</math>. Squaring and rearranging yields:
+Let <math>\sqrt{(9k)^2 - 504} = x</math>. Squaring and rearranging yields:
-<math>(9a)^2 - x^2 = 504</math>
+<math>(9k)^2 - x^2 = 504</math>
-<math>(9a+x)(9a-x) = 504</math>.
+<math>(9k+x)(9k-x) = 504</math>.
-In order for both <math>x</math> and <math>a</math> to be an integer, <math>9a + x</math> and <math>9a - x</math> must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let <math>9a + x = 2m</math> and <math>9a - x = 2n</math>.
+In order for both <math>x</math> and <math>a</math> to be an integer, <math>9k + x</math> and <math>9k - x</math> must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let <math>9k + x = 2m</math> and <math>9k - x = 2n</math>.
 Then:

2007 AMC 12B (Problems • Answer Key • Resources)
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