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24 Nov 2005, 02:20

If k^4 is divisible by 32, which of the following could be the remainder when k is divided by 32?

(1). 2
(2). 4
(3). 6

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24 Nov 2005, 02:27

if k = 2, k^4 = 16 (not divisible by 32)
if k = 4, k^4 = 256 (divisible 32)
if k = 6, k^4 = 1296 (not divisible by 32)

Thus only 4 can be the remainder.
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24 Nov 2005, 02:31

32=2^5 so if 8^4=2^12=2^5x2^5x2^2 is divided by 2^5 it will have reminder 2^2 or 4

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24 Nov 2005, 02:36

gamjatang wrote:

if k = 2, k^4 = 16 (not divisible by 32)
if k = 4, k^4 = 256 (divisible 32)
if k = 6, k^4 = 1296 (not divisible by 32)

Thus only 4 can be the remainder.

agreed. 4 should be the answer.
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Re: Remainder [#permalink]

24 Nov 2005, 02:36

getzgetzu wrote:

If k^4 is divisible by 32, which of the following could be the remainder when k is divided by 32?

(1). 2
(2). 4
(3). 6

let k= 32m + r ( r is the remainder)
k^4= ( 32m+r)^4 ...we know if spreading this expresion we'll get only one term which is not a multiple of 32m ...that is r^4
since k^4 is divisible by 32 ----> r^4 must be divisible by 32
try those choices provided...only 4 satisfies.
4 is correct.

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24 Nov 2005, 02:43

Laxi your responses are stunning! You make things just way too simple! I am absolutely overawed. And Thank You so much.....

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24 Nov 2005, 02:46

getzgetzu wrote:

Laxi your responses are stunning! You make things just way too simple! I am absolutely overawed. And Thank You so much.....

getzgetzu/ You may join us. We have a fan club for Laxi. :lol:

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24 Nov 2005, 02:50

And you are no less great than her gamjatang......

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24 Nov 2005, 02:54

getzgetzu wrote:

And you are no less great than her gamjatang......

yes,o bba is great :wink:

[#permalink] 24 Nov 2005, 02:54

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