Difference between revisions of "1987 AIME Problems/Problem 5"
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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 19:14, 15 January 2021
Contents
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, .
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 |
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