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What is the positive integer n ? (1) For every positive [#permalink]
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21 Sep 2008, 09:21
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What is the positive integer \(n\)? (1) For every positive integer \(m\), \((m+n)!/(m1)!\) is divisible by \(16\) (2) \(n^2  9n + 20 = 0\) Please explain your answer because I couldn't even understand how to approach this problem. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
Last edited by tarek99 on 21 Sep 2008, 14:40, edited 1 time in total.



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 09:30
IMO .... E
1) INSF , m and n can be many different numbers that could be multiple of 16
2) INSF, (n5)(n4)  n could be 5 or 4
Together  INSF  Sat m =2, plug 5 or 4 into equation and both yield integers
INSF... answer E



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 10:08
tarek99 wrote: What is the positive integer \(n\)?
(1) For every integer \(m\), \((m+n)!/(m1)!\) is divisible by \(16\)
(2) \(n^2  9n + 20 = 0\)
Please explain your answer because I couldn't even understand how to approach this problem. (1) Insufficient \((m+n)!/(m1)! = (m+n)*(m+n1)*(m+n2)*...*m\) This expression is divisible by \(2^4\), meaning that when you prime factorize the expression, 2 is a factor at least 4 times. This is the case for many values of m and n. (2) Insufficient \(n^2  9n + 20 = 0\) \((n5)(n4) = 0\) \(n = 5 or 4\) (1) and (2) Insufficient For n=5 or n=4, there are values of m that would allow \((m+n)!/(m1)!\) to be divisible by 16.



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 10:17
The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement: (m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient.
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Re: DS: Factorial [#permalink]
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21 Sep 2008, 10:25
IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. nicely explained +1 u deserve the highest kudos: posts ratio!
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Re: DS: Factorial [#permalink]
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21 Sep 2008, 10:57
i believe factorials only exist for positive numbers..
having said that worst case scenario is that m=1
in that case..n has to be 5..or greater ..cause we dont have enough even numbers..
therefore i will go with C..



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 12:40
IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. Agree, Ian. That'll teach me to pay attention!



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 14:33
Well, the OA is C. Also, the question doesn't mention whether m is positive. I just double checked with the question and it's correct the way it is. I got this from a pdf file, so i'm not sure how completely correct is the question, but this is what it says at least.



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 14:56
IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. that's a good explanation! +1 but what about (m1)!? you avoided it in your explanation. Would you please explain how (m1)! should be be handled? thanks



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Re: DS: Factorial [#permalink]
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21 Sep 2008, 18:13
tarek99 wrote: IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. that's a good explanation! +1 but what about (m1)!? you avoided it in your explanation. Would you please explain how (m1)! should be be handled? thanks Good explanation and good question!!!
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Re: DS: Factorial [#permalink]
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21 Sep 2008, 22:28
IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. Lets say m=5 n=4 9!/4! = 9 X 8 X 7 X 6 X 5. this is divisible by 16 correct How can n not be 4?



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Re: DS: Factorial [#permalink]
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22 Sep 2008, 11:38
icandy wrote: IanStewart wrote: The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:
(m+n)!/(m1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16 take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient. Lets say m=5 n=4 9!/4! = 9 X 8 X 7 X 6 X 5. this is divisible by 16 correct How can n not be 4? Statement 1 says that (m+n)!/(m1)! is divisible by 16 for *every* positive integer m, not just for some particular value of m like m=5. Try m = 3, and you'll see why n cannot be 4. Tarek, I'm not sure I understand your question about (m1)!. In the fraction (m+n)!/(m1)!, the (m1)! cancels with much of (m+n)! to leave us with (m+n)*(m+n1)*...*(m+1)*m. Hope that answers your question. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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