n^4 is divisible by 32

abhi332 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


[spoiler]OA:B[/spoiler]

n^4 is divisible by 2^5
n^4 will be divisible by 2^8 also

n will be divisible by 4

[spoiler]If n is divisible by 4 it can leave only multiples of 4 as the remainder when divided by 32
In the list only B is a multiple of 4 hence B[/spoiler]

hmm.
can you explain how you got to n is divisible by 4?

n^4 is divisible by 2^5. Or, in other words, n^4 has at least five 2s in its prime factorization.

Since the question asks about n, let's think about what this means for n's divisibility by 2:

If integer n were divisible by only one 2, then we know that n^4 would be divisible by 2^4 but not by 2^5. So, in order for n^4 to be divisible by 2^5, n itself must have at least two 2s in its prime factorization. In other words, we can deduce that n is a multiple of 4.

Now, the question becomes: what could the remainder be when we divide a multiple of 4 by 32?

At this point, either we could use the rule that the poster above outlined or we can pick numbers:

32 is a multiple of 4, and 32 divided by 32 leaves a remainder of 0, which doesn't appear in the answer choices. So, the next multiple of 4, namely 36, when divided by 32 leaves a remainder of 4. So the remainder COULD be 4. Since the question is asking for what the remainder COULD be, and since there can only be one correct answer, choose B.

Hi Testluv,

n is the multiple of 4, so n can be 4, 8 16....

if suppose n is 8 and I divide it by 32 will leave remainder 8 only.

I may be missing something..

abhi332 wrote:Hi Testluv,

n is the multiple of 4, so n can be 4, 8 16....

if suppose n is 8 and I divide it by 32 will leave remainder 8 only.

I may be missing something..

Abhi, do 8,12,16 etc feature the answers?

Ok, got it , only 4 is in the answer choice which is multiple of 4..