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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 19 Jan 2008, 14:04
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If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 14:10
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D

the product of the first 8 positive integers is

P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7

2^6=64

1. only 2^6 works. suff.

2. only 2^6 works. suff.
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 14:18
walker wrote:
D

the product of the first 8 positive integers is

P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7

2^6=64

1. only 2^6 works. suff.

2. only 2^6 works. suff.


Big walker, answer is B according to OA. Plus 1 for answering anyways though (can you take another crack at it for me).
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 14:24
dominion wrote:
If the integers A and N are greater than 1, and the product of the first 8 positive integers is a multiple of A^N, what is the value of A?

s1: A^N=64
s2: n=6

Thanks!



S1: A can be 2,4,8 thus n is 6,3,2, 8! is a multiple of any of these combinations.

S2: we have n=6. 8! has 8*7*6*5*4*3*2 ---> 2^7*3^2*5*7 n MUST be 2 or 8! won't be a multiple or be divisible by A^6.

Try 4^6 --> 2^12, this doesnt work. 3^6, there aren't enough 3's to cover this.


Thus s2 is suff.


B
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 14:55
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 ?
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Re: GMATPrep Problem [#permalink] New post 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.
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 21:31
walker wrote:
D

the product of the first 8 positive integers is

P=1*2*3*4*5*6*7*8=2*3*2^2*5*(3*2)*7*2^3=2^6*3^2*5*7

2^6=64

1. only 2^6 works. suff.

2. only 2^6 works. suff.


All fine in your approach,except that you miss other possibilities in ( i ), which can be;

8^2, 4^3 etc. Thus it's "not only" 2^6 works. suff.

But, if n=6 is fixed as in statement ( ii ), then 2^6*3^2*5*7, it is clear that A = 2.

Thus answer is "B"
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Re: GMATPrep Problem [#permalink] New post 19 Jan 2008, 23:12
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I agree :?
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How to solve this? [#permalink] New post 03 Jun 2010, 23:22
How should we solve this?
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Re: How to solve this? [#permalink] New post 04 Jun 2010, 03:14
dimitri92 wrote:
How should we solve this?


We know 8! = 2^6 * 3^2 *5 *7
From question its given a^n *k = 8!

From St-1: a^n*K = 64. a can be 2,4 or 8given a value of n is greater than 1.

From St-2: a^6*K = 8! -> a can only be 2.

Hence B

Last edited by cipher on 04 Jun 2010, 07:40, edited 1 time in total.
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Re: How to solve this? [#permalink] New post 04 Jun 2010, 07:03
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dimitri92 wrote:
How should we solve this?


If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?

Prime factorization would be the best way for such kind of questions.

Given: a^n*k=8!=2^7*3^2*5*7. Q: a=?

(1) a^n=64=2^6=4^3=8^2, so a can be 2, 4, or 8. Not sufficient.

(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.
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Re: Product of integers [#permalink] New post 17 Sep 2010, 00:17
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Geronimo wrote:
If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

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


Prime factorization would be the best way for such kind of questions.

Given: a^n*k=8!=2^7*3^2*5*7. Q: a=?

(1) a^n=64=2^6=4^3=8^2, so a can be 2, 4, or 8. Not sufficient.

(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.
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Re: How to solve this? [#permalink] New post 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] New post 16 Mar 2011, 08:53
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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

Look into this for more on factorials;
http://gmatclub.com/forum/everything-about-factorials-on-the-gmat-85592.html#p641395
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Re: How to solve this? [#permalink] New post 16 Mar 2011, 19:14
From this we have three possibilities for a^n -> 2^7, 3^2 and 4^3 so a can be 2, 3 or 4

From (1) a^n = 64 => a^n = 2^6 or 4^3, so not sufficient

(2) n = 6, so we can rule out 3 or 4, as 2 is the only number with exponent > 6

Hence the answer is B.
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Re: How to solve this? [#permalink] New post 16 Mar 2011, 19:48
I think 3 is ruled out isn't it? all the factorials greater than 5 are Even multiple of 10

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Re: How to solve this? [#permalink] New post 16 Mar 2011, 21:36
In (1) 3 raised to any integer can't be 64. In (2), 3 is ruled out because power/exponent of 3 < 6.
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Re: How to solve this? [#permalink] New post 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

Look into this for more on factorials;
http://gmatclub.com/forum/everything-about-factorials-on-the-gmat-85592.html#p641395



HI FLUKE,
thank you for your help. +1
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Re: a raised to n [#permalink] New post 25 May 2011, 02:15
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:

8! = 8*7*6*5*4*3*2 = (2^3)(7)(2*3)(5)(2^2)(3)(2) = (2^7)(3^2)(5)(7)

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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Re: a raised to n   [#permalink] 25 May 2011, 02:15
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