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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 14 May 2008, 07:26
If the integers a and n are greater than 1 and the product of the first eight positive integers is a multiple of a^n, what is the value of a?

1. a^n=64
2. n=6

I understand why 1. is insufficient: a^n could equal 2^5 or 4^3. However, I'm not certain how knowing n=6 tells me what a must be, given that 8! is a multiple of a^n.

Your help is greatly appreciated.
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Re: Integers a and n [#permalink] New post 14 May 2008, 08:01
MeddlingKid wrote:
If the integers a and n are greater than 1 and the product of the first eight positive integers is a multiple of a^n, what is the value of a?

1. a^n=64
2. n=6

I understand why 1. is insufficient: a^n could equal 2^5 or 4^3. However, I'm not certain how knowing n=6 tells me what a must be, given that 8! is a multiple of a^n.

Your help is greatly appreciated.



statement 1: a^n = 2^6 or 4^3 or 8 ^2 insuff
statement 2: a ^6, if you break down 8!, 8! = 2^7 * 3 * 5 * 7, the only integer that has an exponent of 6 or greater is number 2, suff

B is my guess
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Re: Integers a and n [#permalink] New post 14 May 2008, 08:51
OA is B.

Thank you for the explanation!
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Re: Integers a and n [#permalink] New post 14 May 2008, 11:14
8! = 2^5 * 7 * 3^2 * 5.

The power of 2 is 5 which is less than 6. Please explain

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Re: Integers a and n [#permalink] New post 14 May 2008, 11:28
tekno9000 wrote:
8! = 2^5 * 7 * 3^2 * 5.

The power of 2 is 5 which is less than 6. Please explain


8! = 8X7X6X5X4X3X2X1

8 = 2 x 2 x 2
6 = 2 x 3
4 = 2 x 2
2 = 2

n=6, therefore the multiple must have AT LEAST six 2's. 8! has 7.
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Re: Integers a and n [#permalink] New post 14 May 2008, 13:04
tekno9000 wrote:
8! = 2^5 * 7 * 3^2 * 5.

The power of 2 is 5 which is less than 6. Please explain


You just wrote it down wrong.

8! = 2^7 * 3^2 * 5 * 7
Re: Integers a and n   [#permalink] 14 May 2008, 13:04
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