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# GMATprep-a

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VP
Joined: 03 Apr 2007
Posts: 1291

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26 Jul 2008, 13:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is trick to solve this question?

Attachment:

sheetal_a.GIF [ 20.72 KiB | Viewed 702 times ]
Director
Joined: 10 Sep 2007
Posts: 923

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26 Jul 2008, 13:38
Product of first 8 positive number = 1*2*3*2*2*2*5*2*3*7*2*2*2 = 2^7*3^2*5*7
Told this is multiple of a^n.
Given a>1, and n>1

1) a^n = 64 (2^6 or 4^3 or 8^2). There is differing values so cannot be what is value of a or n.
2) n = 6. There is only possibility to get a^6, that a=2 because no other number repeats more that twice. So the answer.
SVP
Joined: 29 Aug 2007
Posts: 2457

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26 Jul 2008, 13:48
goalsnr wrote:
What is trick to solve this question?
Attachment:
sheetal_a.GIF

a^n (x) = 8!
i. a^n = 64
a^n could be 2^6, 4^4 and 8^2. so insuff..

ii. in 8!, only 2 comes 6 times. so suff.

B.
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Re: GMATprep-a   [#permalink] 26 Jul 2008, 13:48
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# GMATprep-a

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