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Difference between revisions of "1991 AIME Problems/Problem 1"

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== Problem ==
 
== Problem ==
 
Find <math>x^2+y^2_{}</math> if <math>x_{}^{}</math> and <math>y_{}^{}</math> are positive integers such that
 
Find <math>x^2+y^2_{}</math> if <math>x_{}^{}</math> and <math>y_{}^{}</math> are positive integers such that
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<center><math>xy_{}^{}+x+y = 71</math></center>
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<div style="text-align:center"><math>xy_{}^{}+x+y = 71</math></div>
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<center><math>x^2y+xy^2 = 880^{}_{}.</math></center>
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<div style="text-align:center"><math>x^2y+xy^2 = 880^{}_{}.</math></div>
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Define <math>a = x + y</math> and <math>b = xy</math>. Then <math>a + b = 71</math> and <math>ab = 880</math>. Solving these two equations yields a [[quadratic equation|quadratic]]: <math>\displaystyle a^2 - 71a + 880 = 0</math>, which [[factor]]s to <math>\displaystyle (a - 16)(a - 55) = 0</math>. Either <math>a = 16</math> and <math>b = 55</math> or <math>a = 55</math> and <math>b = 16</math>. For the first case, it is easy to see that <math>(x,y)</math> can be <math>(5,11)</math> (or vice versa). In the second case, since all factors of <math>16</math> must be <math>\le 16</math>, no two factors of <math>16</math> can sum greater than <math>32</math>, and so there are no integral solutions for <math>(x,y)</math>. The solution is <math>5^2 + 11^2 = 146</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|before=First question|num-a=2}}
 
{{AIME box|year=1991|before=First question|num-a=2}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 17:56, 11 March 2007

Problem

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that

$xy_{}^{}+x+y = 71$
$x^2y+xy^2 = 880^{}_{}.$

Solution

Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $\displaystyle a^2 - 71a + 880 = 0$, which factors to $\displaystyle (a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$, no two factors of $16$ can sum greater than $32$, and so there are no integral solutions for $(x,y)$. The solution is $5^2 + 11^2 = 146$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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