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

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

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Updated on: 23 Oct 2013, 23:44
10
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Question Stats:

68% (01:31) correct 32% (01:41) wrong based on 1015 sessions

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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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Originally posted by mustdoit on 03 Mar 2010, 00:35.
Last edited by Bunuel on 23 Oct 2013, 23:44, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Re: n^4 is divisible by 32. Find remainder for n/32  [#permalink]

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03 Mar 2010, 12:51
19
24
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]

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03 Mar 2010, 06:39
1
6
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.
##### General Discussion
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Re: n^4 is divisible by 32. Find remainder for n/32  [#permalink]

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

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03 Mar 2010, 07:33
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]

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03 Mar 2010, 08:32
2
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]

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03 Mar 2010, 23:15
Bunuel!! Thanks for the explanation.
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Re: n^4 is divisible by 32. Find remainder for n/32  [#permalink]

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04 Mar 2010, 06:03
1
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]

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27 Aug 2010, 09:50
1
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]

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03 Sep 2010, 20:48
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]

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03 Sep 2010, 21:58
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]

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04 Sep 2010, 23:00
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]

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19 Apr 2014, 21:18
1
1
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

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

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08 Jul 2015, 11:54
1
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]

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18 Jul 2016, 02:52
option 1

Remainder 2 means 32+2=34
n^4= 34^4= 34*34*34*34/32 is not an integer

option 2

Remainder 4 means 32+4=36
n^4= 36*36*36*36/32 is an integer means n^4 is divisible by 32 if we take n=4

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

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15 Oct 2016, 06:34
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 $$4$$.

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.

Hey Bunuel

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

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15 Oct 2016, 06:56
IMO
the number n should be a multiple of 2 greater than 2 itself (2^2, 2^3, 2^4, etc.) ; not necessarily of multiple of 4.

So, we try the first option : 2^2 = 4

4/32 = 0 + remainder 4

So, B is the right answer (4 COULD be a remainder of 4/32).
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If n is an integer and n^4 is divisible by 32, which of the  [#permalink]

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15 Jul 2017, 10:53
This is how I did it.

We're given that $$n^4 ≡ 0 (mod 32)$$

So right off the bat, we know $$n^4$$ is congruent to 32, 64, 96...

We could solve this, but maybe the fastest way is just to plug in the answer choices.

Let's try "2."

$$2^4$$ = $$4^2$$ = 16, and $$16 ≢ 0 (mod 32)$$ This isn't our answer.

Let's try "4."

$$2^4$$ = $$16^2$$ = 256, and $$256 ≡ 0 (mod 32)$$

(because 256 is evenly divisible by 32.)

So "n" could be 4.

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

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25 Jul 2018, 01:24
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 $$4$$.

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]

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25 Jul 2018, 01:28
1
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 $$4$$.

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 &nbs [#permalink] 25 Jul 2018, 01:28

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