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{ | + | 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>. |
==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 11:46, 12 November 2019
Problem
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
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 
.
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. 
