Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"
Geometry285 (talk | contribs) m |
Tigerzhang (talk | contribs) (→Solution) |
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Case 4: <math>x+y=5, xy = 0</math> | Case 4: <math>x+y=5, xy = 0</math> | ||
− | The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - | + | The solutions for <math>x</math> and <math>y</math> are the roots of <math>x^2 - 5x</math>, which are <math>0</math> and <math>5</math>. |
Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>. | Therefore, the answer is <math>1 + 3 + 0 + 5 = 9</math>. |
Revision as of 12:32, 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