Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 13"
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Case 2: <math>x+y=3, xy=4</math>. | Case 2: <math>x+y=3, xy=4</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 - 3x + 4</math>, which are not real. |
Case 3: <math>x+y=4,xy=3</math>. | Case 3: <math>x+y=4,xy=3</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 - 4x + 3</math>, which are <math>1</math> and <math>3</math>. |
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>. | ||
Line 28: | Line 28: | ||
~Revised and Edited by Mathdreams | ~Revised and Edited by Mathdreams | ||
+ | |||
+ | ~Some Edits by BakedPotato66 | ||
==Solution 2== | ==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> | + | Note we are dealing with Pythagorean triples, so <math>xy=\{0,3,4,5\}</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> | <math>\linebreak</math> | ||
~Geometry285 | ~Geometry285 | ||
+ | |||
+ | |||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 16:15, 12 July 2021
Contents
[hide]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
~Some Edits by BakedPotato66
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
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.