Bunuel wrote:
If k is an odd integer, which of the following must be an even integer?
A. k² – 4
B. 3k + 2
C. 2k + 1
D. 12k/8
E. 6k/3
Kudos for correct solution.
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