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product of 8

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Intern
Joined: 13 Apr 2009
Posts: 12

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04 May 2009, 02:26
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can anyone explain this?

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Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS

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04 May 2009, 03:37
LOL.. I still remember this qs.. look break down the first 8 integers (actually 7 cuz, 1 is useless).. so u get
2x3x4x5x6x7x8 = (2^7) x (3^2) x 7 x 5... okay. do that its good for practice. Now since a^n is a factor of the product of these numbers, you can get deduce 'a' if u know 'n'... when n=6, it means a^6, so can u get one of the numbers to become (any #)^6, well turns out you can do that with '2' which is therefore your 'a' meaning the answer is B.
Intern
Joined: 13 Apr 2009
Posts: 12

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04 May 2009, 04:01
Thanks!
So i guess if st2 was one of the following : n=3 or n=4 or n=5 we would still get B as the right answer
cause the only opition a=2
but if st2 was n=2 it would have been E cause a=2 or a=3
right?
Current Student
Joined: 13 May 2008
Posts: 141
Schools: LBS

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04 May 2009, 04:07
aharonm wrote:
Thanks!
So i guess if st2 was one of the following : n=3 or n=4 or n=5 we would still get B as the right answer
cause the only opition a=2
but if st2 was n=2 it would have been E cause a=2 or a=3
right?

Actually if St2 was n=2, St1 would come into play and suggest that a=8 which would mean C.
If n=3, or n=4 or n=5 B would still be the right answer but in each case 'a' would be different, but even so St2 would be alone sufficient.

still I am not very sure with the possibilities u outlined.. do the math and check, maybe i did something wrong
Intern
Joined: 13 Apr 2009
Posts: 12

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04 May 2009, 04:17
why would a be different if n=3 n=4 or n=5?

it is still has to be a=2 because our product is (2^7) x (3^2) x 7 x 5 and the power of "2" is the only one above "2"
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1345

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04 May 2009, 06:51
aharonm wrote:
Thanks!
So i guess if st2 was one of the following : n=3 or n=4 or n=5 we would still get B as the right answer
cause the only opition a=2
but if st2 was n=2 it would have been E cause a=2 or a=3
right?

That's not quite right:

8! = (2^7)(3^2)(5^1)(7^1)

If n=6, 5 or 4, then a must be 2. If, however, n=3, then we have two possible values for a. a could again certainly be equal to 2, but a could also be equal to 4, since 4^3 = (2^2)^3 = 2^6, which is a divisor of 8!. If n = 2, there are quite a few possibilities: a could be 2, 3, 4, 6, 8, 12 or 24.

Of course as soon as you combine the statements, knowing that a is positive, you can definitely solve for a without even reading the rest of the information in the question. That's an immediate clue that C won't likely be the answer here - why would they bother including the information about 8! if it wasn't important in the solution of the problem?
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Manager
Joined: 14 Nov 2008
Posts: 182
Schools: Stanford...Wait, I will come!!!

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05 May 2009, 00:21
Yeah, You are correct.
As our number is (2^7) x (3^2) x 7 x 5
So, if n> 2, we have only one number ..which is 2.
But if it is 2, then we have options 2&3.

aharonm wrote:
why would a be different if n=3 n=4 or n=5?

it is still has to be a=2 because our product is (2^7) x (3^2) x 7 x 5 and the power of "2" is the only one above "2"
Senior Manager
Joined: 07 Jan 2008
Posts: 370

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05 May 2009, 05:31
Stm1:
a^n = 64 = 8^2=4^3=2^6
While 8! = (2^7) x (3^2) x 5 x 7 = (8^2)x(2) x (3^2) x 5 x 7 = (4^3)x(2) x (3^2) x 5 x 7 = (2^6)x(2) x (3^2) x 5 x 7
---> Insufficient!

Stm2:
n=6 ==> Only 2^6 meets the requirement (Because 8! is multiple of a^n) ==> Sufficient.

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Re: product of 8   [#permalink] 05 May 2009, 05:31
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