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# Integers a and n are greater than 1, and the product of the

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Integers a and n are greater than 1, and the product of the [#permalink]  17 Oct 2009, 05:57
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Could anyone help me solve this one?
Integers a and n are greater than 1, and the product of the 1st 8 positive integers is a multiple of a^n. What is the value of a?

A) a^n = 64.
B) n = 6.

The answer was b. I'm confused, particularly since it says: the product of the 1st 8 positive integers, but it does not say what sequence the first 8 integers relates to.

Kusum.
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Joined: 31 Aug 2009
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Re: DS Question. [#permalink]  17 Oct 2009, 07:22
The first 8 positive integers multiplied means 1x2x3x4x5x6x7x8
This equals $$2^7*3^2*5*7$$. For simplicity lets call this number X.
So the question stem says the above number is a multiple of a^n.

Statement 1)
a^n = 64
64 can be 2^6 or 4^3. Both of these are factors of the first X (as they are the same number).
The question asks what is a. A could be both 2 or 4.
Not sufficient.

Statement 2)
This says n=6. If you look at X, the only possible exponent that could be a factor of X is 2^6. This means a=2. The other prime factors to the power of 6 cannot possibly be factors.

Hence ANS = B. Hope that helps.
Re: DS Question.   [#permalink] 17 Oct 2009, 07:22
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# Integers a and n are greater than 1, and the product of the

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