Difference between revisions of "1987 AIME Problems/Problem 5"
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== Problem == | == Problem == | ||
| + | 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 == | ||
| + | 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> | ||
| + | |||
| + | <math>507</math> is equal to <math>3 \cdot 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>. | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/ba6w1OhXqOQ?t=4699 | ||
| + | ~ pi_is_3.14 | ||
| − | == Solution == | + | == Video Solution == |
| + | https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=3704 - AMBRIGGS | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1987|num-b=4|num-a=6}} | |
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | [[Category:Intermediate Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:44, 21 January 2023
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:
![\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]](http://latex.artofproblemsolving.com/e/1/6/e16ed5492ca750a02f850907a11c15b3315706d5.png)
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 by OmegaLearn
https://youtu.be/ba6w1OhXqOQ?t=4699 ~ pi_is_3.14
Video Solution
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. 
