Difference between revisions of "1991 AIME Problems/Problem 1"
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=== Solution 3 === | === Solution 3 === | ||
+ | |||
+ | Let <math>a=x+y</math>, <math>b=xy</math> then we get the equations | ||
+ | <cmath>\begin{align*} | ||
+ | a+b&=71\ | ||
+ | ab&=880 | ||
+ | \end{align*}</cmath> | ||
+ | After getting the [[prime factorization]] of <math></math>880=2^4\cdot5\cdot11, it's easy to obtain the solutions <math>(a,b)=(16,55)</math>. Thus | ||
+ | <cmath>x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}</cmath> | ||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Revision as of 20:55, 4 September 2019
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.
Solution 3
Let ,
then we get the equations
After getting the prime factorization of $$ (Error compiling LaTeX. Unknown error_msg)880=2^4\cdot5\cdot11, it's easy to obtain the solutions
. Thus
~ Nafer
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.