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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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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]
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19 Jan 2008, 15:10
Dthe 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]
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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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Re: GMATPrep Problem [#permalink]
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19 Jan 2008, 15:24
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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]
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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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Re: GMATPrep Problem [#permalink]
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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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Re: GMATPrep Problem [#permalink]
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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. Thus answer is "B"



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Re: GMATPrep Problem [#permalink]
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20 Jan 2008, 00:12
I agree
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How to solve this? [#permalink]
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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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04 Jun 2010, 04:14
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 St1: \(a^n*K = 64\). \(a\) can be 2,4 or 8given a value of \(n\) is greater than 1. From St2: \(a^6*K = 8!\) > a can only be 2. Hence B
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
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]
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17 Sep 2010, 01:17
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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]
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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. 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]
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16 Mar 2011, 09:53
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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; \((2^2)^3*2=(4)^3*2\) Look into this for more on factorials; http://gmatclub.com/forum/everythingaboutfactorialsonthegmat85592.html#p641395
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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 Hence the answer is B.
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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: How to solve this? [#permalink]
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20 Mar 2011, 00:48
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/everythingaboutfactorialsonthegmat85592.html#p641395HI FLUKE, thank you for your help. +1
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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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