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Difference between revisions of "1987 AIME Problems/Problem 5"

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<cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath>
 
<cmath>(3x^2 + 1)(y^2 - 10) = 517 - 10</cmath>
  
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<math>507</math> is equal to <math>3 * 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 = \pm 2</math>. This leaves <math>y^2 - 10 = 39</math>, so <math>y = \pm 7</math>. Thus, <math>3x^2 y^2 = 3 * 4 * 49 = \boxed{588}</math>.
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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>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Revision as of 18:57, 15 June 2019

Problem

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

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 $3 \cdot 13^2$. 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 $x^2 = 4$. This leaves $y^2 - 10 = 39$, so $y^2 = 49$. Thus, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$.

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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