Difference between revisions of "2014 AMC 8 Problems/Problem 13"

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<math>\textbf{(A) }</math> <math>n</math> and <math>m</math> are even <math>\qquad\textbf{(B) }</math> <math>n</math> and <math>m</math> are odd <math>\qquad\textbf{(C) }</math> <math>n+m</math> is even <math>\qquad\textbf{(D) }</math> <math>n+m</math> is odd <math>\qquad \textbf{(E) }</math> none of these are impossible
 
<math>\textbf{(A) }</math> <math>n</math> and <math>m</math> are even <math>\qquad\textbf{(B) }</math> <math>n</math> and <math>m</math> are odd <math>\qquad\textbf{(C) }</math> <math>n+m</math> is even <math>\qquad\textbf{(D) }</math> <math>n+m</math> is odd <math>\qquad \textbf{(E) }</math> none of these are impossible
 
==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{C}</math>.
+
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{D}</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 03:07, 17 January 2020

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

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 $\boxed{D}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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