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When k^4 is divided by 32, the remainder is 0. Which of the following
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Bunuel
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When k^4 is divided by 32, the remainder is 0. Which of the following [#permalink] New post  17 Jan 2021, 23:32
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k is a positive integer. When k^4 is divided by 32, the remainder is 0. Which of the following could be the remainder when k is divided by 32?

I. 2
II. 4
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. II and III only
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rocky620
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Re: When k^4 is divided by 32, the remainder is 0. Which of the following [#permalink] New post  18 Jan 2021, 04:53
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32 = 2^5

k^4=2^8 *n^4 (where n is a integer)
k=2^2*n = 4*n

when k will be divided by 32 the remainder will either be a multiple of 4 or zero (as 32 is also a multiple of 4)

So only ii = 4 is possible

Option B
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wishmasterdj
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Re: When k^4 is divided by 32, the remainder is 0. Which of the following [#permalink] New post  25 Feb 2021, 20:21
chetan2u please explain the approach for this
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chetan2u
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Re: When k^4 is divided by 32, the remainder is 0. Which of the following [#permalink] New post  25 Feb 2021, 20:59
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Bunuel wrote:
k is a positive integer. When k^4 is divided by 32, the remainder is 0. Which of the following could be the remainder when k is divided by 32?

I. 2
II. 4
III. 6

A. I only
B. II only
C. III only
D. I and II only
E. II and III only


wishmasterdj

\(k^4\) is divisible by 32, so \(k^4\) must contain 32 in it.
Distribute all prime factors of 32 in terms of power of 4 as k is to the power of 4.

\(k^4=32y=2^5y=2^4*2y\)
Now k should have one of each of base on the right side \(2^4*2*y\), so 2, 2 and y.
That is k=2*2*y, where y is any positive integer.

If k=4y, then when ever we divide by 32, k will give us a remainder as a multiple of 4.

For example
k=4*1, remainder =4
k=4*2, remainder =8=4*2
k=4*8, remainder =0=4*0
k=4*9, remainder =36=32+4=4

2 and 6 are not multiples of 4. Hence only II

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Re: When k^4 is divided by 32, the remainder is 0. Which of the following [#permalink]
25 Feb 2021, 20:59
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