Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"
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+ | ==Solution 2== | ||
+ | Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4 \}</math>, and we have <math>x+y</math> is a member of the set too. We see <math>x+y=4</math> has <math>x=\{1,3 \}</math> work, but <math>x+y=4</math> has nothing work. If <math>x+y=0</math>, we have <math>x=\{0,5 \}</math> work. The answer is <math>0+1+3+5=\boxed{9}</math> | ||
+ | <math>\linebreak</math> | ||
+ | ~Geometry285 |
Revision as of 11:32, 11 July 2021
Problem
Let and
be nonnegative integers such that
Find the sum of all possible values of
Solution
Notice that since and
are both integers,
and
are also both integers. We can then use casework to determine all possible values of
:
Case 1: .
The solutions for and
are the roots of
, which are not real.
Case 2: .
The solutions for and
are the roots of
, which are not real.
Case 3: .
The solutions for and
are the roots of
, which are
and
.
Case 4:
The solutions for and
are the roots of
, which are
and
.
Therefore, the answer is .
~kante314
~Revised and Edited by Mathdreams
Solution 2
Note we are dealing with Pythagorean triples, so , and we have
is a member of the set too. We see
has
work, but
has nothing work. If
, we have
work. The answer is
~Geometry285