if n^4 is divisible by 32,
n=32a+b we should find b,
so n^4=(32a+b)^4=32(Ax+B)+b^4=> is divisble by 32,
from this b^4 is divisible by 32

2^4-- not divisble
4^4-- divisible
5^4-- not divisible
6^4-- not divisble
10^4-- not divisible

B is the answer

RadiumBall wrote:If n is an integer and n^4 is divisible by 32, which of the following could be the remainder when n is divided by 32?

(A) 2
(B) 4
(C) 5
(D) 6
(E) 10

I'd probably attack this question by picking numbers.

We know that n^4 is divisible by 32, so let's write out some perfect 4th powers, looking for one that's a multiple of 32.

1^4 = 4
2^4 = 16
3^4 = 81
4^4 = 256 = 32*8

So, n could = 4.

What's do you get when you divide 4 by 32? A quotient of 0 and a remainder of 4... choose (B)!

Now, you may be thinking "we got lucky that it was the 4th one, what if it had been 9^4 or something crazy!" The good news is that, on the GMAT, things always work out nicely! So, unless you're convinced that picking numbers is going to be way too time intensive, it often offers a great alternative to complicated algebra or application of concepts.