Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution 1)
Line 9: Line 9:
 
==Solution==
 
==Solution==
 
=== Solution 1 ===
 
=== Solution 1 ===
−
Getting common denominators, we have to find coprime <math>(a,b)</math> such that <math>9ab|9a^2+14b^2</math>. Clearly, <math>3|b</math>. 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, 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>. Clearly, <math>3|b</math>. 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>.
 +
 
 
=== Solution 2 ===
 
=== Solution 2 ===
 
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>.
 
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>.

Revision as of 09:47, 30 July 2015

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

Getting common denominators, we have to find coprime $(a,b)$ such that $9ab|9a^2+14b^2$. Clearly, $3|b$. 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 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last question
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10B_Problems/Problem_25&oldid=71395"