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If the integer a and n are greater than 1 and the product of

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Director
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New post 12 Mar 2007, 05:11
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If the integer a and n are greater than 1 and the product of the first 8 positive intergers is a multiple of a^n, what is the value of a?

1) a^n=64
2) n=6
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New post 12 Mar 2007, 06:07
(C) for me :)

a^n = k * 8! where k is an integer.
and
a > 1 ; n > 1

a = ?

From 1
a^n = 64 = 8 * 8 = 8^2
or
a^n = 64 = 4 * 4 * 4 = 4^3

Thus, a could be, for examples, equal to 8 or 4.

INSUFF.

From 2
n=6

so,
a^6 = k * 8!

For instance, we could choose:
a = 8!
or
a = (2 * 8!)

INSUFF.

Both (1) and (2)
a^6 = 64

Then, a = 2.

SUFF.
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New post 12 Mar 2007, 06:31
That is what i thougth also :)

However, C is not the answer.... Could there be a mistake in Gmatprep?
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New post 12 Mar 2007, 06:46
SimaQ wrote:
If the integer a and n are greater than 1 and the product of the first 8 positive intergers is a multiple of a^n, what is the value of a?

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


It should be B. With n=6, you can find no integer (except a=2) so that a^6 becomes factor of 8!
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Re: DS [#permalink]

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New post 12 Mar 2007, 07:02
singalnitin1 wrote:
SimaQ wrote:
If the integer a and n are greater than 1 and the product of the first 8 positive intergers is a multiple of a^n, what is the value of a?

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


It should be B. With n=6, you can find no integer (except a=2) so that a^6 becomes factor of 8!


Well :)... Perhaps i'm confused by the wording of this question.

When I read "the product of the first 8 positive intergers is a multiple of a^n", it means to me:
a^n = k * 8! where k is an integer.

And not,
8! = k * a^n where k is an integer.

:)
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New post 12 Mar 2007, 09:48
Answer should be B... since I is insufficient avoid A and D.

now from 2 its clear on 2^6= 64 is the multiple of 40320 as it divide it by 630. no other number with power 6 divide it... so a=2 and the answer is B

regards,

Amardeep
  [#permalink] 12 Mar 2007, 09:48
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