1987 AIME Problems/Problem 5

Revision as of 14:37, 11 February 2007 by Azjps (talk | contribs) (solution)

Problem

Find $\displaystyle 3x^2 y^2$ if $\displaystyle x$ and $\displaystyle y$ are integers such that $\displaystyle y^2 + 3x^2 y^2 = 30x^2 + 517$.

Solution

If we move the $x^2$ term to the left side, it is factorable:

$(3x^2 + 1)(y^2 - 10) = 517 - 10$

$507$ is equal to $3 * 13^2$. Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. 169 doesn't work either, so $3x^2 + 1 = 13$, and $x = \pm 2$. This leaves $y^2 - 10 = 39$, so $y = \pm 7$. Thus, $\displaystyle 3x^2 y^2 = 3 * 4 * 49 = 588$.

See also

1987 AIME (ProblemsAnswer KeyResources)
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