Bunuel wrote:
k is a positive integer. When k^4 is divided by 32, the remainder is 0. Which of the following could be the remainder when k is divided by 32?
I. 2
II. 4
III. 6
A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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