When k^4 is divided by 32, the remainder is 0. Which of the following

k^4 divided by 32 leaves a remainder of 0, can express as \(k^4 = 32n = 2^5n\) for some integer n.

Taking the fourth root of this equation, we get \(k = 2\sqrt[4]{2n}\).

Given that k is an integer and \(k = 2\sqrt[4]{2n}\), it follows that \(\sqrt[4]{2n}\) must also be an integer. Therefore, k can assume values like:

4 for \(n = 2^3\);
8 for \(n = 2^3*2^4\);
12 for \(n = 2^3*3^4\);
20 for \(n = 2^3*5^4\);
24 for \(n = 2^3*2^4*3^4\);
28 for \(n = 2^3*7^4\);
32 for \(n = 2^3*2^8\) and so on.

Essentially, k will always be a multiple of 4. Consequently, when dividing k by 32, the possible remainders are 4, 8, 12, 16, 20, 24, 28, or 0 – all multiples of 4.

Answer: B.