32 = 2^5

k^4=2^8 *n^4 (where n is a integer)
k=2^2*n = 4*n

when k will be divided by 32 the remainder will either be a multiple of 4 or zero (as 32 is also a multiple of 4)

So only ii = 4 is possible

Option B

chetan2u please explain the approach for this

wishmasterdj

\(k^4\) is divisible by 32, so \(k^4\) must contain 32 in it.
Distribute all prime factors of 32 in terms of power of 4 as k is to the power of 4.

\(k^4=32y=2^5y=2^4*2y\)
Now k should have one of each of base on the right side \(2^4*2*y\), so 2, 2 and y.
That is k=2*2*y, where y is any positive integer.

If k=4y, then when ever we divide by 32, k will give us a remainder as a multiple of 4.

For example
k=4*1, remainder =4
k=4*2, remainder =8=4*2
k=4*8, remainder =0=4*0
k=4*9, remainder =36=32+4=4

2 and 6 are not multiples of 4. Hence only II

B