Strictly speaking, the above approaches are backwards. The question does not say "If the remainder is 2 when k is divided by 32, is k^4 divisible by 32?", which is the question the above solutions answer. Fortunately it doesn't make a difference to the answer here, but it will on other questions.
If k^4 is divisible by 32, which of the following could be the remainder when k is divided by 32?
I. 2
II. 4
III. 6
If you know k^4 is divisible by 32, then k^4 is divisible by 2^5. What must k be divisible by? If k were divisible by 2 but *not* by 2^2, then k^4 would only be divisible by 2^4 = 16. So k must be divisible by at least 2^2 = 4. Thus, k is a multiple of 4, and since 32 is also a multiple of 4, when k is divided by 32, the remainder will be a multiple of 4. If that is unclear, algebraically, we know k = 4a for some a. When we divide k by 32, we can write, with q as the quotient, and r as the remainder:
k = 32q + r
4*a = 32q + r
r = 4*a - 32q
r = 4(a - 8q)
so r must be a multiple of 4. Only II is possible.
I'd note that if you were going to do the problem 'backwards', anju has chosen the best numbers here; in a remainder problem, if you are going to pick numbers, you will always get the smallest (and therefore easiest) set of numbers by letting the quotient be zero, (and not one). If the remainder is 2 when k is divided by 32, k could certainly be equal to 2.
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