2007 AMC 12B Problems/Problem 24: Difference between revisions

Revision as of 22:19, 16 November 2013

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 $\mathrm {(A)}$

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 subtractions, because if we had addition signs, that means $a$ is 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 divisible 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 $\mathrm {(A)}$

@@ Line 21: / Line 21: @@
 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>
+==Solution 2==
+Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m}</math> We get--
+<math>9a^2 - 9mab + 14b^2 = 0</math>
+Factoring this, we get <math>4</math> equations-
+<math>(3a-2b)(3a-7b) = 0</math>
+<math>(3a-b)(3a-14b) = 0</math>
+<math>(a-2b)(9a-7b) = 0</math>
+<math>(a-b)(9a-14b) = 0</math>
+(It's all subtractions, because if we had addition signs, that means <math>a</math> is the opposite sign of <math>b</math>)
+Now we look at these, and see that-
+<math>3a=2b</math>
+<math>3a=b</math>
+<math>3a=7b</math>
+<math>3a=14b</math>
+<math>a=2b</math>
+<math>9a=7b</math>
+<math>a=b</math>
+<math>9a=14b</math>
+This gives us <math>8</math> solutions, but we note that the middle term needs to give you back <math>9m</math>.
+For example, in the case
+<math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not divisible by <math>9m</math> for whatever integar <math>m</math>.
+Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math>
 ==See Also==
 {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

2007 AMC 12B (Problems • Answer Key • Resources)
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2007 AMC 12B Problems/Problem 24: Difference between revisions

Revision as of 22:19, 16 November 2013

Contents

Problem 24

Solution

Solution 2

See Also