1987 AIME Problems/Problem 5: Difference between revisions
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<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>. | <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 == | |||
https://youtu.be/ba6w1OhXqOQ?t=4699 | |||
~ pi_is_3.14 | |||
== See also == | == See also == | ||
Revision as of 18:14, 15 January 2021
Problem
Find 
if 
and 
are integers such that 
.
Solution
If we move the 
term to the left side, it is factorable:
![\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]](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
https://youtu.be/ba6w1OhXqOQ?t=4699 ~ pi_is_3.14
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
