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When the integer k is divided by 12, the remainder is 3.

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When the integer k is divided by 12, the remainder is 3. [#permalink] New post 19 Jul 2008, 02:44
When the integer k is divided by 12, the remainder is 3. Which of the following, when divided by 18, will have a remainder of 6 ?
I. 2k
II. 6k
III. 4k + 6

(A) I only
(B) II only
(C) III only
(D) None
(E) I, II, and III
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Re: Remainder Problem [#permalink] New post 19 Jul 2008, 03:05
I think the answer is D.

if m is a running integer then k = 12m+3

I. 2k = 24m + 6,
II. 6k = 72m + 18
III. 4k + 6 = 48m + 18

When divided by 18, none of them will give a remainder of 6.
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Re: Remainder Problem [#permalink] New post 19 Jul 2008, 07:18
Yes, I also did the same break down... however, I wasn't sure how to find out the remainder for statement I and statement III.

Statement II is easy ---
II. 6k = 72m + 18

Both 72m and 18 are fully divisible by 18--- thus giving NO remainder.

But I'm not sure about I or III where 24 and 48 do not divide into 18 evenly.

Any takers??

I'll put the OA in a little bit when more people respond.
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Re: Remainder Problem [#permalink] New post 19 Jul 2008, 08:06
sirlogic wrote:
When the integer k is divided by 12, the remainder is 3. Which of the following, when divided by 18, will (read must) have a remainder of 6 ?
I. 2k
II. 6k
III. 4k + 6

(A) I only
(B) II only
(C) III only
(D) None
(E) I, II, and III

possible valus of k: 3, 15, 27, 39, 51,....

take any to disprove and you are fine as they are looking for must response.
I. 15*2 = 30; remainder is not 6
II. 6*3 = 18; remainder is not 6
II 4k+6 4*3+6=18; remainder is not 6.
None.
D
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Re: Remainder Problem [#permalink] New post 19 Jul 2008, 08:11
For stmt 1 and 3, the remainder varies with m (as used in the previous post)
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Re: Remainder Problem [#permalink] New post 19 Jul 2008, 08:12
Very good. Thanks.

OA is indeed D.
Re: Remainder Problem   [#permalink] 19 Jul 2008, 08:12
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When the integer k is divided by 12, the remainder is 3.

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