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 | ||
| − | <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> | ||
| + | __TOC__ | ||
== Solution == | == Solution == | ||
| + | === Solution 1 === | ||
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>. | 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>. | ||
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| + | === Solution 2 === | ||
| + | Since <math>xy + x + y + 1 = 72</math>, this can be factored to <math>(x + 1)(y + 1) = 72</math>. As <math>x</math> and <math>y</math> are [[integer]]s, the possible sets for <math>(x,y)</math> (ignoring cases where <math>x > y</math> since it is symmetrical) are <math>(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)</math>. The second equation factors to <math>(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11</math>. The only set with a factor of <math>11</math> is <math>(5,11)</math>, and checking shows that it is our solution. | ||
== See also == | == See also == | ||
Revision as of 19:33, 12 March 2007
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 | ||
