Difference between revisions of "1987 AIME Problems/Problem 5"
(→Video Solution) |
m (→Solution) |
||
Line 2: | Line 2: | ||
Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are [[integer]]s such that <math>y^2 + 3x^2 y^2 = 30x^2 + 517</math>. | Find <math>3x^2 y^2</math> if <math>x</math> and <math>y</math> are [[integer]]s such that <math>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 [[SFFT| | + | If we move the <math>x^2</math> term to the left side, it is factorable with [[SFFT|Simon's Favorite Factoring Trick]]: |
<cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath> | <cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath> |
Revision as of 15:36, 19 September 2022
Contents
[hide]Problem
Find if and are integers such that .
Solution
If we move the term to the left side, it is factorable with Simon's Favorite Factoring Trick:
is equal to . Since and are integers, cannot equal a multiple of three. doesn't work either, so , and . This leaves , so . Thus, .
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=4699 ~ pi_is_3.14
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=3704 - AMBRIGGS
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.