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If n is an integer and n^4 is divisible by 32, which of the

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If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   Updated on: 24 Oct 2013, 00:44
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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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B

Originally posted by mustdoit on 03 Mar 2010, 01:35.
Last edited by Bunuel on 24 Oct 2013, 00:44, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   03 Mar 2010, 13:51
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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

\(n^4\) being divisible by 32 implies \(n^4=32k=2^5k\), for some integer k.

Taking the fourth root of this equation gives: \(n=2\sqrt[4]{2k}\).

Since n is an integer, the least value of n (\(n_{min}\)) is obtained when k = 8. Hence, \(n_{min}=2\sqrt[4]{2*8}=4\).

4 divided by 32 gives a 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 \(k = 2^3 * 3^4\), and so on. Basically, n will be a multiple of 4. So \(\frac{n}{32}\) can give us the remainder of 4, 8, 12, 16, 20, 24, 28, 0 - all multiples of 4. The only multiple of 4 in the answer choices is 4.


Answer: B.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   03 Mar 2010, 07:39
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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.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   03 Mar 2010, 09:32
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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....

32 = 2^5
n^4/32 is an integer

n^4 is even and at the minimum (n^4 and n) should be a multiple of 4.

so the min value of n is 4 and remainder is 4

B
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   04 Mar 2010, 00:15
Bunuel!! Thanks for the explanation.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   04 Mar 2010, 07:03
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IMO B

Take \(n^4 = 2^5 * K\) where K can be of form \(K = 2^{4m-5}\)

=> \(n = 2^m\)

now m can be any integer greater than 1
( m cannot be 1 as K is an integer thus, 4m-5 > 0 )

now m=2,3,4....so on gives n =4, 8 , 16........36, 40... so on

now 4 can be the remainder when n=36.

Since we are dividing by 32, remainder must be multiple of 4 as n is multiple of 4.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   27 Aug 2010, 10:50
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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.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   03 Sep 2010, 21:48
Gurpreet,
Can you please explain why does 32A be of the form 2^(4z)? Thanks
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   03 Sep 2010, 22:58
mainhoon wrote:
Gurpreet,
Can you please explain why does 32A be of the form 2^(4z)? Thanks


32*A = n^4 and n is integer. take 4th root on both the sides.

Both LHS and RHS should have integral values. This is only possible when 32A = 2^{4z}
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink] New post   05 Sep 2010, 00:00
Always use method explained by bunuel above in this type of the problems
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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   19 Apr 2014, 22:18
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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
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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   08 Jul 2015, 12:54
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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
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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   25 Jul 2018, 02:24
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.



Bunuel, can you please explain how from here \(n^4=2^5k\) you got ths \(n=2\sqrt[4]{2k}\) ?
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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   25 Jul 2018, 02:28
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dave13 wrote:
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.



Bunuel, can you please explain how from here \(n^4=2^5k\) you got ths \(n=2\sqrt[4]{2k}\) ?


By taking the fourth root:
\(n^4=2^5k\);

\(n=\sqrt[4]{2^5k}=\sqrt[4]{2^4*2*k}=2\sqrt[4]{2k}\).
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Re: If n is an integer and n^4 is divisible by 32, which of the [#permalink] New post   18 Nov 2023, 02:32
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