Difference between revisions of "2014 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
− | Since <math>n^2+m^2</math> is even, either both <math>n^2</math> and <math>m^2</math> are even, or they are both odd. Therefore, <math>n</math> and <math>m</math> are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, <math>n+m</math> must be even. The answer, then, is <math>\boxed{ | + | Since <math>n^2+m^2</math> is even, either both <math>n^2</math> and <math>m^2</math> are even, or they are both odd. Therefore, <math>n</math> and <math>m</math> are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result, <math>n+m</math> must be even. The answer, then, is <math>\boxed{C}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=12|num-a=14}} | {{AMC8 box|year=2014|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:23, 12 November 2022
Contents
[hide]Problem
If and
are integers and
is even, which of the following is impossible?
and
are even
and
are odd
is even
is odd
none of these are impossible
Video Solution
https://www.youtube.com/watch?v=boXUIcEcAno
https://youtu.be/_3n4f0v6B7I ~savannahsolver
Solution
Since is even, either both
and
are even, or they are both odd. Therefore,
and
are either both even or both odd, since the square of an even number is even and the square of an odd number is odd. As a result,
must be even. The answer, then, is
.
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.