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

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==Solution 2==
 
==Solution 2==
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Instead of using logic to solve this, we can just plug in random numbers. If <math>n^{2}+m^{2}</math> is even, they are both even or both odd. So just use numbers like 1 and 3. 1+3=4, and 4 is even so the answer is <math>\boxed{C}</math>.
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Instead of using logic to solve this, we can just plug in random numbers. Since <math>2^2+4^2 = 20</math> which is even, we see that it is possible for both <math>m</math> and <math>n</math> to be even, and for <math>m+n</math> to be even.  <math>\boxed{C}</math>.
  
 
~Trex226
 
~Trex226
 
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*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
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~awesomeguy856
  
 
==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 00:08, 23 November 2022

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

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. AMC logo.png

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