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Re: GMATPrep Problem [#permalink]
19 Jan 2008, 15:53

pmenon wrote:

Hey gmatblackbelt, Im not following the logic above for statement 2.

We know that A^6 = (2^7*3^2*5)k ... where do we go from here ?

A^6 does not equal 2^7*3^2*5*7

8! is divisble by A^6 the only way that this could be is for A to equal 2. It cannot equal 1 since the main stem said A and N are bothg greater than 1.

Re: How to solve this? [#permalink]
15 Mar 2011, 22:49

Bunuel wrote:

dimitri92 wrote:

How should we solve this?

(2) n=6 --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence a=2. Sufficient.

Answer: B.

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?
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Re: How to solve this? [#permalink]
16 Mar 2011, 08:53

1

This post received KUDOS

MICKEYXITIN wrote:

Bunuel wrote:

dimitri92 wrote:

How should we solve this?

(2) n=6 --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence a=2. Sufficient.

Answer: B.

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; (2^2)^3*2=(4)^3*2

Re: How to solve this? [#permalink]
19 Mar 2011, 23:48

fluke wrote:

MICKEYXITIN wrote:

Bunuel wrote:

Hi bunuel, Thank you for your instruction. I have another query: how can we eliminate the possibility that: 8! is not a multiple of 4^6 (which means a#4) quickly?

8! has 2^7 as its factor. Thus maximum exponent of 4 is 3; (2^2)^3*2=(4)^3*2

When looking at Statement 1, we know a^n = 64 and a and n are positive integers greater than 1, so a could be 2, 4 or 8 (since 2^6 = 4^3 = 8^2 = 64). The information in the stem isn't actually important here.

When we look at Statement 2, we know that 8! is divisible by a^6. If we prime factorize 8!, we find:

We need this prime factorization to be divisible by a^6 where a > 1. Looking at the prime factorization, the only possibility is that a = 2 (since for any other prime p besides 2, we can't divide the factorization above by p^6, nor can we divide the factorization above by (2^2)^6 = 2^12).
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