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

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== Problem ==
 
== Problem ==
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Find <math>\displaystyle 3x^2 y^2</math> if <math>\displaystyle x</math> and <math>\displaystyle y</math> are integers such that <math>\displaystyle y^2 + 3x^2 y^2 = 30x^2 + 517</math>.
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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 factorable:
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If we move the <math>x^2</math> term to the left side, it is factorable with [[SFFT|Simon's Favorite Factoring Trick]]:
  
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:<math>(3x^2 + 1)(y^2 - 10) = 517 - 10</math>
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<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 [[integer]]s, <math>3x^2 + 1</math> cannot equal a multiple of three. 169 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>\displaystyle 3x^2 y^2 = 3 * 4 * 49 = 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>.
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== Video Solution by OmegaLearn ==
 +
https://youtu.be/ba6w1OhXqOQ?t=4699
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~ pi_is_3.14
 +
 
 +
== 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}}
 
{{AIME box|year=1987|num-b=4|num-a=6}}
 +
 +
[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]
 +
{{MAA Notice}}

Latest revision as of 03:44, 21 January 2023

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 with Simon's Favorite Factoring Trick:

\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]

$507$ 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, $3x^2 y^2 = 3 \times 4 \times 49 = \boxed{588}$.

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

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