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

Revision as of 13:11, 24 November 2023

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 (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 $n^2+m^2$ $\boxed{C}$ .(Typo in the answer)

This solution(above not by me) is incorrect, because the answer is $\boxed{D}$ . If we replace $n$ and $m$ with odd numbers, the value outputs to a even same with inputting with even numbers. This crosses out A and B. It also crosses out C because odd + odd is even. For D, we have to have one odd and one even. If we take 1 and 2 for an example - 1+4=5 which is odd. We can also see that odd * odd is odd, and even * even is even, and odd + even is odd. This crosses out E, and we get $\boxed{D}$ .

-anonymous(Top solution) - Multpi12(bottom solution)

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.

@@ Line 20: / Line 20: @@
 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>n^2+m^2</math> <math>\boxed{C}</math>.(Typo in the answer)
-This solution(above not by me) is incorrect as the answer is <math>\boxed{D}</math>. If we replace <math>n</math> and <math>m</math> with odd numbers, the value outputs to a even same with inputting with even numbers. This crosses out A and B. It also crosses out C because odd + odd is even. For D, we have to have one odd and one even. If we take 1 and 2 for an example - 1+4=5 which is odd. We can also see that odd * odd is odd, and even * even is even, and odd + even is odd. This crosses out E, and we get <math>\boxed{D}</math>.
+This solution(above not by me) is incorrect, because the answer is <math>\boxed{D}</math>. If we replace <math>n</math> and <math>m</math> with odd numbers, the value outputs to a even same with inputting with even numbers. This crosses out A and B. It also crosses out C because odd + odd is even. For D, we have to have one odd and one even. If we take 1 and 2 for an example - 1+4=5 which is odd. We can also see that odd * odd is odd, and even * even is even, and odd + even is odd. This crosses out E, and we get <math>\boxed{D}</math>.
 -anonymous(Top solution)

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