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# Absolute values

Author Message
Director
Joined: 22 Aug 2007
Posts: 566

Kudos [?]: 71 [0], given: 0

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03 Sep 2007, 01:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

ds
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Kudos [?]: 71 [0], given: 0

SVP
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 167 [0], given: 0

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03 Sep 2007, 02:04
(C) for me

|x| < 4
<=> -4 < x < 4

From (1)
x is divisable by 3
<=> x = 3*k where k is an integer
=> x=-3 or x=0 or x=3

INSUFF.

From (2)
x is divisable by 2
<=> x = 2*j where j is an integer
=> x=-2 or x=0 or x=2

INSUFF.

Both (1) & (2)
Only 0 could satisfy both statments.

SUFF.

Kudos [?]: 167 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5034

Kudos [?]: 438 [0], given: 0

Location: Singapore

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03 Sep 2007, 02:12
St1:
x can be 0,-3 or 3. Insufficient.

St2:
x can be 0,-2 or 2. Insufficient.

Using both st1 and st2:
x must be 0. The remaining integers divisible by both 2 and 3 are multiples of 6, which falls out of the boundary set by |x| < 4.

Ans C

Kudos [?]: 438 [0], given: 0

03 Sep 2007, 02:12
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