Difference between revisions of "1987 AIME Problems/Problem 5"
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Find <math>\displaystyle 3x^2 y^2</math> if <math>\displaystyle x</math> and <math>\displaystyle y</math> are integers such that <math>\displaystyle y^2 + 3x^2 y^2 = 30x^2 + 517</math>. | Find <math>\displaystyle 3x^2 y^2</math> if <math>\displaystyle x</math> and <math>\displaystyle y</math> are integers such that <math>\displaystyle y^2 + 3x^2 y^2 = 30x^2 + 517</math>. | ||
== Solution == | == Solution == | ||
| − | + | If we move the <math>x^2</math> term to the left side, it is factorable: | |
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| + | :<math>(3x^2 + 1)(y^2 - 10) = 517 - 10</math> | ||
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| + | <math>507</math> is equal to <math>3 * 13^2</math>. Since <math>x</math> and <math>y</math> are [[integer]]s, <math>3x^2 + 1</math> cannot equal a multiple of three. 169 doesn't work either, so <math>3x^2 + 1 = 13</math>, and <math>x = \pm 2</math>. This leaves <math>y^2 - 10 = 39</math>, so <math>y = \pm 7</math>. Thus, <math>\displaystyle 3x^2 y^2 = 3 * 4 * 49 = 588</math>. | ||
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== See also == | == See also == | ||
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{{AIME box|year=1987|num-b=4|num-a=6}} | {{AIME box|year=1987|num-b=4|num-a=6}} | ||
Revision as of 13:37, 11 February 2007
Problem
Find 
if 
and 
are integers such that 
.
Solution
If we move the 
term to the left side, it is factorable:
is equal to 
. Since 
and 
are integers, 
cannot equal a multiple of three. 169 doesn't work either, so 
, and 
. This leaves 
, so 
. Thus, 
.
See also
| 1987 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
