Problem Solving

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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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

A lot of Integer Properties questions can be answered quickly by testing values that satisfy the given information.

So, for example, if n² - kn is EVEN, it COULD be the case that n = 1 and k = 1
Now use these values to check the answer choices
A) n² = 1² = 1 (not even - ELIMINATE)
B) k² = 1² = 1 (not even - ELIMINATE)
C) 2n + k² = 2(1) + 1² = 3 (not even - ELIMINATE)
D) n(k + 1) = 1(1 + 1) = 2 (EVEN - keep!!)
E) k(k + n) = 1(1 + 1) = 2 (EVEN - keep!!)

We are now down to answer choices D and E.
So let's test a second pair of values.
If n² - kn is EVEN, it could also be the case that n = 2 and k = 1
Now use these values to check the remaining answer choices
D) n(k + 1) = 2(1 + 1) = 4 (EVEN - keep!!)
E) k(k + n) = 1(1 + 2) = 3 (not even - ELIMINATE)

By the process of elimination the answer must be D

Cheers,
Brent

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

If we choose n = 3 and k = 1, we see that n^2 = 9 and k^2 = 1 are odd; thus, we can eliminate answer choices A and B.

If we choose n = 2 and k = 1, we see that 2n + k^2 = 4 + 1 = 5 and k(k + n) = 1(1 + 2) = 3 are odd; thus, we eliminate answer choices C and E as well.

The only remaining answer choice is D.

Answer: D