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1991 AIME Problems/Problem 1: Difference between revisions

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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
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Find <math>x^2+y^2_{}</math> if <math>x_{}^{}</math> and <math>y_{}^{}</math> are positive integers such that
<div style="text-align:center;"><math>xy_{}^{}+x+y = 71</math></div>
<div style="text-align:center;"><math>xy_{}^{}+x+y = 71</math></div>
<div style="text-align:center"><math>x^2y+xy^2 = 880^{}_{}.</math></div>
<div style="text-align:center"><math>x^2y+xy^2 = 880^{}_{}.</math></div><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>


__TOC__
__TOC__
== Solution ==
== Solution ==
=== Solution 1 ===
=== Solution 1 ===

Revision as of 14:43, 18 November 2015

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

Solution 1

Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$, which factors to $(a - 16)(a - 55) = 0$. 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 $5^2 + 11^2 = 146$.

Solution 2

Since $xy + x + y + 1 = 72$, this can be factored to $(x + 1)(y + 1) = 72$. As $x$ and $y$ are integers, the possible sets for $(x,y)$ (ignoring cases where $x > y$ since it is symmetrical) are $(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)$. The second equation factors to $(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11$. The only set with a factor of $11$ is $(5,11)$, and checking shows that it is our solution.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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