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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]
03 Mar 2010, 06:39

2

This post was BOOKMARKED

Given: n^4/32 = int n^4/2^5 = int 1/2(n/2)^4 = int

try n= int = 2,4,6,8... becuase n/2 must be an integer

n=2 wont work because 1/2(n/2)^4 needs to be an integer. n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) n=8, result = 1/2(8/2)^4 = 1/2(4)^4 = 1/2x16x16= 16x8, so the remainder of n/32 = remainder of 16x8/32 = 4 (option B)

Once you understand that the question is asking 1/2(n/2)^4 must be an integer then you can quickly run the numbers and figure out the remainder of n/32.

Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]
03 Mar 2010, 07:10

rao_1857 wrote:

Given: n^4/32 = int n^4/2^5 = int 1/2(n/2)^4 = int

try n= int = 2,4,6,8... becuase n/2 must be an integer

n=2 wont work because 1/2(n/2)^4 needs to be an integer. n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) n=8, result = 1/2(8/2)^4 = 1/2(4)^4 = 1/2x16x16= 16x8, so the remainder of n/32 = remainder of 16x8/32 = 4 (option B)

Once you understand that the question is asking 1/2(n/2)^4 must be an integer then you can quickly run the numbers and figure out the remainder of n/32.

Maybe I'm missing something but why are you calculating the result, and then getting the remainder of the result divided by 32? Shouldn't you be getting the remainder of n divided by 32? For example, using your statement above: n=4, result = 1/2 (2)^4 = 8, so the remainder of n/32 = remainder of 8/32 = 8 (not an option) Result is indeed 8, but you need to divide 4 by 32 (0 remainder 4), not 8 by 32

If its done the way I described above, the answer could be A, D or E

Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]
03 Mar 2010, 12:51

8

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Expert's post

7

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mustdoit wrote:

If n is an integer and n4 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

Solving this would clarify lots of concepts !!! Solution coming....

Given: \(n^4=32k=2^5k\) --> \(n=2\sqrt[4]{2k}\) --> as \(n\) is an integer, the least value of \(n\) is obtained when \(k=8\) --> \(n_{min}=2\sqrt[4]{2*8}=4\) --> \(\frac{n_{min}}{32}=\frac{4}{32}\) gives remainder of \(4\).

Answer: B.

To elaborate more: as \(n\) is an integer and \(n=2\sqrt[4]{2k}\), then \(\sqrt[4]{2k}\) must be an integer. So \(n\) can take the following values: \(4\) for \(k=8\), \(8\) for \(k=2^3*2^4\), \(12\) for \(k=2^3*3^4\)... Basically \(n\) will be multiple of \(4\). So \(\frac{n}{32}\) can give us the reminder of 4, 8, 12, 16, 20, 24, 28, 0 - all multiples of 4. Only multiple of 4 in answer choices is 4. _________________

Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]
27 Aug 2010, 09:50

1

This post received KUDOS

I have made typo mistakes and that makes it bad solution. I dont rem where I was dreaming that day. Since n^4 is divisible by 32 =>

\(n^4 = 32*A\) but for n to be an integer \(32*A\) should be reduced to an integer when we take 4th root on both sides thus 32*A must be of the form \(2^{4z}\) => \(A = 2^{4z-5}\) => n is of the form \(2^z\)

for\(z =1 , 4z-5 <0\) thus not possible. for \(z = 2 , n = 4\)

when 4 is divided by 32 the remainder is 4. _________________

Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]
23 Oct 2013, 21:12

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink]
19 Apr 2014, 21:18

1

This post was BOOKMARKED

Concept Tested :

Remainders

background:

a*b*c/d then

if Rem(a/d) =x, Rem(b/d)=y, Rem(c/d)=z then

x*y*z/d

Sol:

Try numbers using options.

(1). 2

means 32+2 =34

REM(34*34*34*34/32) must be 0

REM(2*2*2*2/32) is 16

(2). 4

means 32+4=36

REM(36*36*36*36/32) must be 0

REM(4*4*4*4/32) is 0

Hence the answer _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink]
29 May 2015, 23:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink]
08 Jul 2015, 11:54

mustdoit 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

Following is how I tackle this problem, please give comment!

32 = 2^5. n^4 = n*n*n*n --> n^4 is divisible by 32 only if n is divisible at least by 4 (because n^4 need one more factor of "2" to be divisible by 2^5) --> The smallest value of n is 4 --> remainder of n : 32 = 4 --> B

gmatclubot

Re: If n is an integer and n^4 is divisible by 32, which of the
[#permalink]
08 Jul 2015, 11:54

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