Remainder, exponents

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Remainder, exponents

by avenus » Sun May 31, 2009 10:53 am
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
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by ssmiles08 » Sun May 31, 2009 11:05 am
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?

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by avenus » Tue Jun 02, 2009 1:08 am
OA D

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by scoobydooby » Tue Jun 02, 2009 8:05 am
i get B too, worked out just like ssmiles08 did.
hey avenus,
do you have the OE? curious how they solved it :)

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by Stuart@KaplanGMAT » Tue Jun 02, 2009 12:10 pm
scoobydooby wrote:i get B too, worked out just like ssmiles08 did.
hey avenus,
do you have the OE? curious how they solved it :)
4^4 is 256, which is certainly divisible by 32.

4/32 is 0rem4, so it's definitely possible to get a remainder of 4. If the OA is 6, the OA is wrong.
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by doclkk » Wed Jun 03, 2009 12:26 pm
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?
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.

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by ssmiles08 » Wed Jun 03, 2009 12:59 pm
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.
If you draw out a long division on the paper: 4 would be inside and 32 is outside.

0 is the quotient which gives you 32*0 = 0; so 4-0 = 4: remainder

Hope that clarifies a little

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by doclkk » Wed Jun 03, 2009 1:09 pm
ssmiles08 wrote:
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.
If you draw out a long division on the paper: 4 would be inside and 32 is outside.

0 is the quotient which gives you 32*0 = 0; so 4-0 = 4: remainder

Hope that clarifies a little
Yea I got it now, just have never done this on a GMAT problem yet ...?