GeminiHeat wrote:
If n and k are integers and n^2 – kn is even, which of the following must be even?
A) n^2
B) k^2
C) 2n + k^2
D) n(k + 1)
E) k(k + n)
n and k are integers.
We have no info about individual terms n and k (even or odd). That eliminates options A and B.
In option C, 2n is even for sure. Again, we have no info about the individual term k - hence eliminated
Option D: n(k + 1) = nk + n
We know that n^2 – kn is even
Note: if A-B is even, A+B is also even
Also: if A-B is odd, A+B is also odd
(Try to think why that is true)
=> Thus, n² + kn is also even
Again, if n is even, n² will be even. Similarly, if n is odd, n² will be odd.
=> Thus: n + kn is also even
[we simply replace n² by n]
Thus, answer is option D.
Option E: k(k + n) = k² + kn
However, we have no individual info about k (for k²) and hence cannot determine even or odd.
Answer D.
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