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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]

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19 Jan 2008, 15:04
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65% (hard)

Question Stats:

62% (02:05) correct 38% (02:23) wrong based on 273 sessions

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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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19 Jan 2008, 15:10
1
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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19 Jan 2008, 15:18
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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19 Jan 2008, 15:24
1
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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19 Jan 2008, 15:55
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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19 Jan 2008, 16:53
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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19 Jan 2008, 22:31
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.

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20 Jan 2008, 00:12
1
I agree
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04 Jun 2010, 00:22
How should we solve this?
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Re: How to solve this?  [#permalink]

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Updated on: 04 Jun 2010, 08:40
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

Originally posted by cipher on 04 Jun 2010, 04:14.
Last edited by cipher on 04 Jun 2010, 08:40, edited 1 time in total.
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Re: How to solve this?  [#permalink]

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04 Jun 2010, 08:03
2
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.

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17 Sep 2010, 01:17
1
1
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.

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Re: How to solve this?  [#permalink]

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15 Mar 2011, 23:49
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.

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]

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16 Mar 2011, 09:53
1
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.

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;
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Re: How to solve this?  [#permalink]

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16 Mar 2011, 20:14
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

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Re: How to solve this?  [#permalink]

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16 Mar 2011, 20:48
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]

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16 Mar 2011, 22:36
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: a raised to n  [#permalink]

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25 May 2011, 03:15
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: If the integers a and n are greater than 1 and the product  [#permalink]

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21 Apr 2015, 06:45
Statement (1) : can be multiple combinations, hence insufficient

Statement (2) : According to this, (8/a) + (8/a^2) + (8/a^3)....... = 6 , where a^n<8
For a = 2
8/2 + 8/4 = 4 + 2 = 6 . Hence , a =2 .. Sufficient
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Re: If the integers a and n are greater than 1 and the product  [#permalink]

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