BTGModeratorVI wrote: ↑Tue Feb 18, 2020 11:10 am
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)
Answer:
D
Source: Manhattan Prep
So, we have n^2 – kn = n(n – k) is even.
Case 1: Say n = even, then (n – k) can be even or odd. Thus, k can be even or odd.
Case 2: Say n = odd, then (n – k) must be even. Since n is odd, we have k = odd.
Let's see each option one by one.
A) n^2: Since n can be even or odd, n^2 can also be even or odd.
B) k^2: From Case 1, k can be even or odd, so k^2 can also be even or odd.
C) 2n + k^2:
Whether n is even or odd, 2n is even. We have already seen in Option B that k^2 can also be even or odd; thus, 2n + k^2 can be even or odd.
D) n(k + 1):
Case 1: If n is even, it is immaterail whether (k + 1) is even or odd; we see that n(k + 1) is even.
Case 2: If n is odd, we know that k is also odd; thus, (k + 1) would be even. Thus, n(k + 1) = Odd*Even = Even.
This option is must be true.
E) k(k + n):
Case 1: If k is even, it is immaterial whether (k + n) is even or odd; we see that k(k + n) is even.
Case 2: If k is odd and n is even, then k(k + n) = Odd*(Odd + Even) = Odd*Odd = Odd.
No unique answer.
The correct answer:
D
Hope this helps!
-Jay
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