Difference between revisions of "2007 AMC 10B Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | 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{ | + | === 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>. | ||
+ | === 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>. | ||
==See Also== | ==See Also== |
Revision as of 10:28, 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:
is an integer?
Contents
Solution
Solution 1
Getting common denominators, we have to find coprime such that . Clearly, . Since and are coprime, . Similarly, . However, cannot be as only has solutions when . Therefore, and . Checking them all, we see that they work and the answer is .
Solution 2
Let . We can then write the given expression as where is an integer. We can rewrite this as a quadratic, . By the Quadratic Formula, . We know that must be rational, so must be a perfect square. Let . Then, . The factors pairs of are and , and , and , and and . Only and and and give integer solutions, and and and , respectively. Plugging these back into the original equation, we get possibilities for , namely and .
See Also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
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All AMC 10 Problems and Solutions |
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