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
(A) 2
(B) 4
(C) 5
(D) 6
(E) 10
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4^4 is 256, which is certainly divisible by 32.scoobydooby wrote:i get B too, worked out just like ssmiles08 did.
hey avenus,
do you have the OE? curious how they solved it

Could you explain this a little more. how does 4/32 provide R 4 ?ssmiles08 wrote:I got 4.
Not sure if it is right but here is my explanation:
if n^4 is divisible by 32, the n^4 must have at least five 2's in it's product.
since n^4 = n^2* n^2
every prime number in n must occur in sets of 2.
2*2*2*2*2 = 32;
32 is 2^5, so three more 2's need to be added to make n have four sets of two 2's.
so the smallest value n could possibly be is 4
4/32, would give the remainder of 4.
I am not sure if I explained it really well. What is the OA?
If you draw out a long division on the paper: 4 would be inside and 32 is outside.doclkk wrote:
Could you explain this a little more. how does 4/32 provide R 4 ?
4/32 = 1/8... how can you have a remainder when the numerator is larger than the denominator.
Yea I got it now, just have never done this on a GMAT problem yet ...?ssmiles08 wrote:If you draw out a long division on the paper: 4 would be inside and 32 is outside.doclkk wrote:
Could you explain this a little more. how does 4/32 provide R 4 ?
4/32 = 1/8... how can you have a remainder when the numerator is larger than the denominator.
0 is the quotient which gives you 32*0 = 0; so 4-0 = 4: remainder
Hope that clarifies a little