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Director
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Joined: 19 Mar 2007
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Pls see attached [#permalink] New post 03 May 2007, 05:36
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Director
Director
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Joined: 19 Mar 2007
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Re: DS - GMATPrep (a b k m) [#permalink] New post 03 May 2007, 05:46
kirakira wrote:
nick_sun wrote:
Pls see attached


(C) it is


well, OA is C
Manager
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 [#permalink] New post 03 May 2007, 06:02
The question is:

Is (b^m/a^k) an integer ?

(1) Insufficient because if m<k it does not work.

(2) Insufficient because we do not know whether "a" is a factor of "b"

Then (1) and (2) sufficient.

Answer C.
VP
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 [#permalink] New post 03 May 2007, 12:34
my way of solving is this: plug in

a=4
b=8

statement 1

4^m
8^k

if we will replace m=2 k=1 then a is bigger
then b and as so can't be a factor of b. but if k=1 m=2 then a is a factor of b.

insufficient

statement 2

a^2
b^1

a,b can take any number so:

insufficient

statement 1&2

we can now know that:

4^1 = 4
8^2 = 64

so the answer is (C)

(*) we can prove that for every a^n and b^(n+1)
  [#permalink] 03 May 2007, 12:34
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