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# K^4 is divisible by 32. When K is divided by 32, what can be

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Senior Manager
Joined: 19 May 2004
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K^4 is divisible by 32. When K is divided by 32, what can be [#permalink]

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11 Jul 2004, 11:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

K^4 is divisible by 32.
When K is divided by 32, what can be the remainder ?

I) 2 II) 4 III) 6

A) I, II
B) I, III
C) II
D) II, III
E) I, II, III

Good luck!

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Director
Joined: 05 May 2004
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Location: San Jose, CA
Re: What can be the remainder ? [#permalink]

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11 Jul 2004, 15:35
I think the Ans is E

let K=32n+R
Then
K^4=(32n+R)^4
The only entity in the above expression that is not divisible by 32 is R^4
For K^4 to be divisible by 32, R^4 has to be divisible by 32
Hence R can be 2,4,6

Waiting for the OA!

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CIO
Joined: 09 Mar 2003
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11 Jul 2004, 23:28
Great question, Dookie!

If K^4 is divisible by 32, then k itself must be a multiple of 4. If it were only a multiple of 2, then raising it to the 4th power wouldn't make it a multiple of 32, which is 2^5.

Since it's a multiple of 4, and 32 is a multiple of 4, the only remainders could be multiples of 4, also.

So only C works.

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Director
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12 Jul 2004, 07:33
concur with (C)

Picking numbers.

For k^4 to be divisible by 32, k must be multiple of 4 such as : 4, 8, 12....

so it will have remainder in multiples of 4. So, (C) survives.

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Senior Manager
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12 Jul 2004, 09:29
OA is C.

Ian, great explanation !

If someone has similiar questions please post them.
On the real test i was attacked by a similiar monster.

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Director
Joined: 05 May 2004
Posts: 574

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Location: San Jose, CA
Re: What can be the remainder ? [#permalink]

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12 Jul 2004, 10:25
srijay007 wrote:
I think the Ans is E

let K=32n+R
Then
K^4=(32n+R)^4
The only entity in the above expression that is not divisible by 32 is R^4
For K^4 to be divisible by 32, R^4 has to be divisible by 32
Hence R can be 2,4,6

Waiting for the OA!

OH WHAT A STUPID MISTAKE I MADE!!!

R^4 has to be divisible by 32 is possible only when R=4
(NOT R=2 or 6); hence C is the answer.

Nice question

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Manager
Joined: 21 Jun 2004
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16 Jul 2004, 23:53
jpv wrote:
concur with (C)

Picking numbers.

For k^4 to be divisible by 32, k must be multiple of 4 such as : 4, 8, 12....

so it will have remainder in multiples of 4. So, (C) survives.

You guys rock..........

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16 Jul 2004, 23:53
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# K^4 is divisible by 32. When K is divided by 32, what can be

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