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# n^4 is divisible by 32. Find remainder for n/32

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n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 01:35
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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

Solving this would clarify lots of concepts !!! Solution coming....
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Last edited by mustdoit on 03 Sep 2010, 23:03, edited 1 time in total.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 06:44
Would be interested in seeing the solution to this one.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 07:39
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.
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  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
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 08:33
2^4=16 so not possible
and 2^5 = 32

so v need (2^2)^2

4 satisfies this condition
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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 09:32
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

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Re: n^4 is divisible by 32. Find remainder for n/32 [#permalink]  03 Mar 2010, 13:51
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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.

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

forz =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]  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]  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]  05 Sep 2010, 00:00
Always use method explained by bunuel above in this type of the problems
Re: n^4 is divisible by 32. Find remainder for n/32   [#permalink] 05 Sep 2010, 00:00
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