2014 AMC 8 Problems/Problem 13

Problem 13

If $n$ and $m$ are integers and $n^2+m^2$ is even, which of the following is impossible?

$\textbf{(A) }$ $n$ and $m$ are even $\qquad\textbf{(B) }$ $n$ and $m$ are odd $\qquad\textbf{(C) }$ $n+m$ is even $\qquad\textbf{(D) }$ $n+m$ is odd $\qquad \textbf{(E) }$ none of these are impossible

Video Solution (CREATIVE THINKING)

https://youtu.be/jQLIxT5qCTY

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=boXUIcEcAno ~David

https://youtu.be/_3n4f0v6B7I ~savannahsolver

Solution

Since $n^2+m^2$ is even, either both $n^2$ and $m^2$ are even, or they are both odd. Therefore, $n$ and $m$ 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, $n+m$ must be even. The answer, then, is $n^2+m^2$ $\boxed{(\text{D})}$ is odd.

Solution 2

We can assign values of n and m as 1,3. Then 1^2 $+ 3^2$ is 10. We can also assign values of n and m to be even, like 2 and 4. After calculating, we can determine that $n^2$ + $m^2$ could be odd or even. Then, we can conclude that n+m is not odd. Hence our answer choice is $\boxed{(\text{D})}$ -TheNerdWhoIsNerdy.

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.

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