Difference between revisions of "2014 AMC 8 Problems/Problem 13"
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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>\boxed{C}</math>. (Typo in the answer) | + | 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>. | ||
| + | |||
| + | -anonymous(Top solution) | ||
| + | - Multpi12(bottom solution) | ||
==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 13:10, 24 November 2023
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
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 
.(Typo in the answer)
This solution(above not by me) is incorrect as the answer is 
. If we replace and 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 .
-anonymous(Top solution) - Multpi12(bottom solution)
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. 
