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

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

7

This post received KUDOS

Expert's post

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

forz =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

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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

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 _____________________________________________________________________________

gmatclubot

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