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 | + | <!-- 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__ | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
Revision as of 14:43, 18 November 2015
Problem
Find 
if 
and 
are positive integers such that
Solution
Solution 1
Define 
and 
. Then 
and 
. Solving these two equations yields a quadratic: 
, which factors to 
. Either 
and 
or 
and 
. For the first case, it is easy to see that 
can be 
(or vice versa). In the second case, since all factors of 
must be 
, no two factors of can sum greater than 
, and so there are no integral solutions for . The solution is 
.
Solution 2
Since 
, this can be factored to 
. As 
and 
are integers, the possible sets for (ignoring cases where 
since it is symmetrical) are 
. The second equation factors to 
. The only set with a factor of 
is , 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
