2014 AMC 8 Problems/Problem 13

Problem

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

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

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 $\boxed{D}$ .

Solution 2

Instead of using logic to solve this, we can just plug in random numbers. Since $2^2+4^2 = 20$ which is even, we see that it is possible for both $m$ and $n$ to be even, and for $m+n$ to be even. $\boxed{C}$ .

~Trex226 .

note that this solution (which gave a wrong answer) involves plugging in random numbers, which is difficult for this particular problem due to answer choice E

~awesomeguy856

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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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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2014 AMC 8 Problems/Problem 13

Contents

Problem

Video Solution

Solution

Solution 2

See Also