Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]

Show Tags

03 Mar 2010, 07:39

5

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]

Show Tags

03 Mar 2010, 08:10

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

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]

Show Tags

27 Aug 2010, 10:50

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]

Show Tags

23 Oct 2013, 22:12

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).

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]

Show Tags

19 Apr 2014, 22:18

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]

Show Tags

30 May 2015, 00:31

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]

Show Tags

08 Jul 2015, 12:54

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

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

Show Tags

12 Jul 2016, 15:20

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.
_________________

If n is an integer and n^4 is divisible by 32, which of the [#permalink]

Show Tags

15 Oct 2016, 07:34

Bunuel wrote:

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.

Thanks for your solution....could you please let me know if this is the right way of solving this?

n^4 is divisible by 32, implies \(\frac{n^4}{2^5}\) clearly, if n=2, then the above condition does not stand true, as we will have one 2 remaining in the denominator, hence n has to be a multiple of at least 4 (because we need another 2 in the Nr)

Chose a value of n which is greater than 32 and multiple of 4. \(\frac{36}{32}\), remainder = 4

OR could we simply say Remainder/Denominator = 4/32?
_________________

If you analyze enough data, you can predict the future.....its calculating probability, nothing more!