Difference between revisions of "2007 AMC 10B Problems/Problem 25"
(added solution) |
|||
Line 7: | Line 7: | ||
<math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math> | <math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math> | ||
+ | ==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>36</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{4 (A)}</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== | ||
{{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}} | {{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}} |
Revision as of 16:25, 21 February 2012
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?
Solution
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 |
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 |