swerve wrote: ↑

Wed Aug 12, 2020 10:14 am

If \(k\) is an odd integer, which of the following must be an even integer?

A. \(k^2-4\)
B. \(3k+2\)
C. \(2k+1\)
D. \(\dfrac{12k}{8}\)
E. \(\dfrac{6k}{3}\)

The OA is E

Source: Magoosh

The key word in must, which means the correct answer will yield an EVEN value for ALL odd values of k.
So, if we find an odd value for k that does NOT yield an even output, then we can eliminate that answer choice.

Let's see what happens when k = 1 (1 is a nice odd integer to work with)
Plug k = 1 to get...
A. 1² – 4 = -3. -3 is NOT even. ELIMINATE A
B. 3(1) + 2 = 5. 5 is NOT even. ELIMINATE B
C. 2(1) + 1 = 3. 3 is NOT even. ELIMINATE C
D. 12(1)/8 = 3/2. 3/2 is NOT even. ELIMINATE D

By the process of elimination, the correct answer must be E

Cheers,
Brent