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Difference between revisions of "2007 AMC 10B Problems/Problem 25"

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For reference, when given two numbers a and b, <math>a|b</math> means that <math>b</math> is divisible by <math>a</math>*
 
For reference, when given two numbers a and b, <math>a|b</math> means that <math>b</math> is divisible by <math>a</math>*
  
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Getting common denominators, we have to find coprime <math>(a,b)</math> such that <math>9ab|(9a^2+14b^2)</math>. b is divisible by 3 because 14 is not a multiple of three in the equation, so b must be so balance it and make them integers. Since <math>a</math> and <math>b</math> are coprime, <math>a|9a^2+14b^2 \implies a|14</math>. Similarly, <math>b|9</math>. However, <math>b</math> cannot be <math>9</math> as <math>81a|81 \cdot 14 + 9a^2</math> only has solutions when <math>3|a</math>. Therefore, <math>b=3</math> and <math>a \in \{1,2,7,14\}</math>. Checking them all (Or noting that <math>4</math> is the smallest answer choice), we see that they work and the answer is <math>\boxed{\mathrm{(A) \ } 4}</math>.
+
Getting common denominators, we have to find coprime <math>(a,b)</math> such that <math>9ab|(9a^2+14b^2)</math>. b is divisible by 3 because 14 is not a multiple of three in the equation, so b must balance it and make them integers. Since <math>a</math> and <math>b</math> are coprime, <math>a|9a^2+14b^2 \implies a|14</math>. Similarly, <math>b|9</math>. However, <math>b</math> cannot be <math>9</math> as <math>81a|81 \cdot 14 + 9a^2</math> only has solutions when <math>3|a</math>. Therefore, <math>b=3</math> and <math>a \in \{1,2,7,14\}</math>. Checking them all (Or noting that <math>4</math> is the smallest answer choice), we see that they work and the answer is <math>\boxed{\mathrm{(A) \ } 4}</math>.
  
  

Revision as of 18:37, 15 December 2019

Problem

How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:

$\frac{a}{b} + \frac{14b}{9a}$

is an integer?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many}$

Solution

Solution 1

For reference, when given two numbers a and b, $a|b$ means that $b$ is divisible by $a$*

Getting common denominators, we have to find coprime $(a,b)$ such that $9ab|(9a^2+14b^2)$. b is divisible by 3 because 14 is not a multiple of three in the equation, so b must balance it and make them integers. Since $a$ and $b$ are coprime, $a|9a^2+14b^2 \implies a|14$. Similarly, $b|9$. However, $b$ cannot be $9$ as $81a|81 \cdot 14 + 9a^2$ only has solutions when $3|a$. Therefore, $b=3$ and $a \in \{1,2,7,14\}$. Checking them all (Or noting that $4$ is the smallest answer choice), we see that they work and the answer is $\boxed{\mathrm{(A) \ } 4}$.


  • as per Wikipedia

Solution 2

Let $x = \frac{a}{b}$. We can then write the given expression as $x+\frac{14}{9x} = k$ where $k$ is an integer. We can rewrite this as a quadratic, $9x^2 - 9kx + 14 = 0$. By the Quadratic Formula, $x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}$. We know that $x$ must be rational, so $9k^2-56$ must be a perfect square. Let $9k^2-56 = n^2$. Then, $56 = 9k^2-n^2 = (3k - n)(3k + n)$. The factors pairs of $56$ are $1$ and $56$, $2$ and $28$, $4$ and $14$, and $7$ and $8$. Only $2$ and $28$ and $4$ and $14$ give integer solutions, $k = 5$ and $n = 13$ and $k = 3$ and $n = 5$, respectively. Plugging these back into the original equation, we get $\boxed{\mathrm{(A) \ } 4}$ possibilities for $x$, namely $\frac{1}{3}, \frac{14}{3}, \frac{2}{3},$ and $\frac{7}{3}$.

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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