Difference between revisions of "1991 AIME Problems/Problem 1"
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=== Solution 4 === | === Solution 4 === | ||
− | From the first equation, we know x+y=71-xy<math>. We factor the second equation as xy(71-xy)=880< | + | From the first equation, we know <math>x+y=71-xy</math>. We factor the second equation as xy(71-xy)=880<math>. Let </math>a=xy$ and rearranging we get a^2-71a+880=(a-16)(a-55)=0. We have two cases: (1) x+y=16 and xy=55 OR (2) x+y=55 and xy=16. We find the former is true for (x,y) = (5,11). x^2+y^2=121+25=146. (sorry, I don't know how to use Latex). |
== See also == | == See also == |
Revision as of 16:43, 23 February 2020
Problem
Find if and are positive integers such that
Contents
[hide]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.
Solution 3
Let , then we get the equations After finding the prime factorization of , it's easy to obtain the solution . Thus Note that if , the answer would exceed which is invalid for an AIME answer. ~ Nafer
Solution 4
From the first equation, we know . We factor the second equation as xy(71-xy)=880a=xy$ and rearranging we get a^2-71a+880=(a-16)(a-55)=0. We have two cases: (1) x+y=16 and xy=55 OR (2) x+y=55 and xy=16. We find the former is true for (x,y) = (5,11). x^2+y^2=121+25=146. (sorry, I don't know how to use Latex).
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.