1991 AIME Problems/Problem 1
Problem
Find if
and
are positive integers such that
![$xy_{}^{}+x+y = 71$](http://latex.artofproblemsolving.com/d/c/5/dc565ff6101912a3697a0d86a8599417d28f166f.png)
![$x^2y+xy^2 = 880^{}_{}.$](http://latex.artofproblemsolving.com/e/2/8/e28b7a0ff087588f697f1eb0f77b6838d6a1dfe2.png)
Contents
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)=880
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
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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