DS - remainder

This is a really cool question.

Answer is C.

Factors of 24 are 2,2,2,3

1) try out a couple odd numbers...
n=3 yields 6... R=6
n=5 yields 24... R =0
So since both are different, INSUFFICIENT

2) try out a couple non-multiple of three numbers:
n=5 yields 24... R=0
n=8 yields 63... R=15
So since both are different, INSUFFICIENT

Together:
n=5 yields 24... R=0
n=7 yields 48... R=0
n=11 yields 120... R=0
n=13 yields 168... R=0
.....
so, SUFFICIENT!

Why?

Since we're multiplying n-1 and n+1, we're multiplying the two adjacent numbers. Whenever n is odd, we'll be multiplying two even numbers together... one is at least divisible by 2^2 while the other is divisible by 2... So we have at least 2x2x2 as factors.. Recall that 2,2,2,3 are the factors of 24. When n is not divisible by 3, we also end up with either n+1 or n-1 being divisible by 3, since picking any 3 adjacent numbers will yield one number that is divisible by three. So under those circumstances, 24 will always be a factor... always leaving a remainder of 0.

Thanks buddy!