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

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Current Student
Joined: 01 Apr 2008
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Schools: Chicago Booth '11
If the integers a and n are greater than 1 and the product [#permalink]

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14 May 2008, 08:26
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 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.

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Director
Joined: 23 Sep 2007
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Re: Integers a and n [#permalink]

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14 May 2008, 09:01
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.

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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Current Student
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Schools: Chicago Booth '11
Re: Integers a and n [#permalink]

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14 May 2008, 09:51
OA is B.

Thank you for the explanation!

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Intern
Joined: 01 Jan 2007
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Re: Integers a and n [#permalink]

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14 May 2008, 12:14
8! = 2^5 * 7 * 3^2 * 5.

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

Regards,

Rocky

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SVP
Joined: 24 Aug 2006
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Re: Integers a and n [#permalink]

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14 May 2008, 12:28
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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Director
Joined: 26 Jul 2007
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Re: Integers a and n [#permalink]

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14 May 2008, 14:04
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

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Re: Integers a and n   [#permalink] 14 May 2008, 14:04
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