Difference between revisions of "1987 AIME Problems/Problem 5"
m (→Solution) |
|||
Line 6: | Line 6: | ||
<cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath> | <cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath> | ||
− | <math>507</math> is equal to <math>3 * 13^2</math>. Since <math>x</math> and <math>y</math> are integers, <math>3x^2 + 1</math> cannot equal a multiple of three. <math>169</math> doesn't work either, so <math>3x^2 + 1 = 13</math>, and <math>x = | + | <math>507</math> is equal to <math>3 * 13^2</math>. Since <math>x</math> and <math>y</math> are integers, <math>3x^2 + 1</math> cannot equal a multiple of three. <math>169</math> doesn't work either, so <math>3x^2 + 1 = 13</math>, and <math>x^2 = 4</math>. This leaves <math>y^2 - 10 = 39</math>, so <math>y^2 = 49</math>. Thus, <math>3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}</math>. |
== See also == | == See also == |
Revision as of 21:39, 16 March 2017
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. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.