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If the integers a and n are greater than 1 and the product

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If the integers a and n are greater than 1 and the product [#permalink] New post 09 Mar 2008, 01:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If the integers a and n are greater than 1 and the product of the first 8 positive
integers is a multiple of a^n, what is the value of a?

1) a^n = 64
2) n = 6
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Re: exponents and integers [#permalink] New post 09 Mar 2008, 06:35
From St-1: \(a^n*K = 8!\). a can be anything 2,3,4 given a value of n. Basically 2 degrees of freedom.

From St-2: \(a^6*K = 8!\) -> a can only be 2.

So B.
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Re: exponents and integers [#permalink] New post 09 Mar 2008, 10:27
yes, I agree with B as well

we need to find the number that has the most number of factors in 8!, it is only 2..and thus n=6
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Re: exponents and integers [#permalink] New post 09 Mar 2008, 10:49
GMAT TIGER wrote:
neelesh wrote:
From St-1: \(a^n*K = 8!\). a can be anything 2,\(3\),4 given a value of n. Basically 2 degrees of freedom.

From St-2: \(a^6*K = 8!\) -> a can only be 2.

So B.

how is a =3?


in statment 1) a can only be 1, 2 and 64 etc but yes not 3..or 5 or 7..
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Re: exponents and integers [#permalink] New post 09 Mar 2008, 12:36
how can we quickly tell that 8! will be divisible by 64 ?
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Re: exponents and integers [#permalink] New post 09 Mar 2008, 20:52
pmenon wrote:
how can we quickly tell that 8! will be divisible by 64 ?


8! = 1x2x3x4x5x6x7x8
8! = 1x2x3x2x2x5x2x3x7x2x2x2

8! has altogather seven 2s.
64 = 2x2x2x2x2x2
64 has 6 2s.

so 8!/64 = 2^7 x 3^2 x 5 x 7 / 2^6 = 2 x 3^2 x 5 x 7 = 630
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Re: exponents and integers   [#permalink] 09 Mar 2008, 20:52
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If the integers a and n are greater than 1 and the product

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