Difference between revisions of "2014 AMC 8 Problems/Problem 13"
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If <math>n</math> and <math>m</math> are integers and <math>n^2+m^2</math> is even, which of the following is impossible? | If <math>n</math> and <math>m</math> are integers and <math>n^2+m^2</math> is even, which of the following is 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 | <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 | ||
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==Video Solution (CREATIVE THINKING)== | ==Video Solution (CREATIVE THINKING)== | ||
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==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>n^2+m^2</math> <math>\boxed{(\text{D})n + m is odd.}</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>n^2+m^2</math> <math>\boxed{(\text{D})}</math> is odd. |
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| + | ==Solution 2== | ||
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| + | We can assign values of n and m as 1,3. Then 1^2<math> + 3^2</math> is 10. We can also assign values of n and m to be even, like 2 and 4. After calculating, we can determine that <math>n^2</math> + <math>m^2</math> could be odd or even. Then, we can conclude that n+m is not odd. Hence our answer choice is <math>\boxed{(\text{D})}</math>-TheNerdWhoIsNerdy. | ||
==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}} | ||
Latest revision as of 12:22, 15 September 2024
Contents
Problem 13
If 
and 
are integers and 
is even, which of the following is impossible?
and are even 
and are odd 
is even 
is odd 
none of these are impossible
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=boXUIcEcAno ~David
https://youtu.be/_3n4f0v6B7I ~savannahsolver
Solution
Since is even, either both 
and 
are even, or they are both odd. Therefore, and 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, must be even. The answer, then, is 
is odd.
Solution 2
We can assign values of n and m as 1,3. Then 1^2
is 10. We can also assign values of n and m to be even, like 2 and 4. After calculating, we can determine that + could be odd or even. Then, we can conclude that n+m is not odd. Hence our answer choice is -TheNerdWhoIsNerdy.
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. 
