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How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:
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==Problem==
  
<math>\mathrm frac {a}/{b}</math> + <math>\mathrm frac {14b}/{9a}</math>  
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How many pairs of positive integers <math>(a,b)</math> are there such that <math>a</math> and <math>b</math> have no common factors greater than <math>1</math> and:
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<cmath>\frac{a}{b} + \frac{14b}{9a}</cmath>
  
 
is an integer?
 
is an integer?
  
(A) 4
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<math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math>
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==Solution==
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=== Solution 1 ===
  
(B) 6
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We will use the divisibility notation (<math>a|b</math>), which means <math>b</math> is divisible by <math>a</math>. Getting common denominators, we have to find coprime <math>(a,b)</math> such that <math>9ab|(9a^2+14b^2)</math>. <math>b</math> is divisible by 3 because 14 is not a multiple of three in the equation, so <math>b</math> 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>.
  
(C) 9
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=== Solution 2 ===
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Let <math>x = \frac{a}{b}</math>. We can then write the given expression as <math>x+\frac{14}{9x} = k</math> where <math>k</math> is an integer. We can rewrite this as a quadratic, <math>9x^2 - 9kx + 14 = 0</math>. By the Quadratic Formula, <math>x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}</math>. We know that <math>x</math> must be rational, so <math>9k^2-56</math> must be a perfect square. Let <math>9k^2-56 = n^2</math>. Then, <math>56 = 9k^2-n^2 = (3k - n)(3k + n)</math>. The factors pairs of <math>56</math> are <math>1</math> and <math>56</math>, <math>2</math> and <math>28</math>, <math>4</math> and <math>14</math>, and <math>7</math> and <math>8</math>. Only <math>2</math> and <math>28</math> and <math>4</math> and <math>14</math> give integer solutions, <math>k = 5</math> and <math>n = 13</math> and <math>k = 3</math> and <math>n = 5</math>, respectively. Plugging these back into the original equation, we get <math>\boxed{\mathrm{(A) \ } 4}</math> possibilities for <math>x</math>, namely <math>\frac{1}{3}, \frac{14}{3}, \frac{2}{3},</math> and <math>\frac{7}{3}</math>.
  
(D) 12
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==See Also==
  
(E) infinitely many
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{{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}}
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{{MAA Notice}}

Revision as of 17:47, 18 September 2020

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

We will use the divisibility notation ($a|b$), which means $b$ 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 $\boxed{\mathrm{(A) \ } 4}$.

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last question
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All AMC 10 Problems and Solutions

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