2014 AMC 8 Problems/Problem 13
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, because 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 |
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