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# Consecutive integers

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Manager
Joined: 15 Oct 2008
Posts: 103

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24 Oct 2008, 15:42
Hi all,
Pls help me with this problem?
List 6 factors of the product of 5 consecutive even intergers?

My answer is: the six factors of this products will be 1 and the 5 consecutive even integers themselves. Is that right?
Ex: the 5 consecutive even integers are: 2, 4, 6, 8, 10.
So the 6 factors of this product will be: 1, 2, 4, 6, 8, 10.

There is a different explanation in the MGMAT - Number Properties. I understand the explanation but want to check to see if my understanding is right or wrong.
Thanks much.

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VP
Joined: 30 Jun 2008
Posts: 1018

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24 Oct 2008, 21:43
nganle08 wrote:
Hi all,
Pls help me with this problem?
List 6 factors of the product of 5 consecutive even integers?

My answer is: the six factors of this products will be 1 and the 5 consecutive even integers themselves. Is that right?
Ex: the 5 consecutive even integers are: 2, 4, 6, 8, 10.
So the 6 factors of this product will be: 1, 2, 4, 6, 8, 10.

There is a different explanation in the MGMAT - Number Properties. I understand the explanation but want to check to see if my understanding is right or wrong.
Thanks much.

Yes you are right. The product of 5 different numbers will have at least the 5 numbers as factors plus the number 1, which is a factor of all numbers
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Senior Manager
Joined: 28 Feb 2007
Posts: 296

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25 Oct 2008, 01:02
Does the question imply anything about which set of the 6 factors.
Because In ur example 2,4,6,8,9 the product is 2^7*3^3 which has (7+1)*(3+1)= 32 factors. Out of these 32 factors we can make 32!/30!*6! sets that have 6 members each.
VP
Joined: 30 Jun 2008
Posts: 1018

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25 Oct 2008, 01:10
UMB wrote:
Does the question imply anything about which set of the 6 factors.
Because In ur example 2,4,6,8,9 the product is 2^7*3^3 which has (7+1)*(3+1)= 32 factors. Out of these 32 factors we can make 32!/30!*6! sets that have 6 members each.

UMB .. the question says list 6 factors... it can be any of the six factors ... so most obviously 5 consecutive even numbers will themselves be the factors, including the number 1, making them 6 factors.

btw, can you explain this bit "Out of these 32 factors we can make 32!/30!*6! sets that have 6 members each."

shouldn't it be 32!/(26!*6!) or am I missing something ??
_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Senior Manager
Joined: 28 Feb 2007
Posts: 296

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25 Oct 2008, 01:53
Amit u r right. SImple calculation error. I couldn;t solve 32-6 =? equation. 26! it is .
Manager
Joined: 15 Oct 2008
Posts: 103

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27 Oct 2008, 10:10
UMB wrote:
Does the question imply anything about which set of the 6 factors.
Because In ur example 2,4,6,8,9 the product is 2^7*3^3 which has (7+1)*(3+1)= 32 factors. Out of these 32 factors we can make 32!/30!*6! sets that have 6 members each.

How can you come up with 9? I did not list 9 in my sample set.
Senior Manager
Joined: 28 Feb 2007
Posts: 296

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27 Oct 2008, 11:24
TYpo. sorry. u indeed not mentioned 9-my bad.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post 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.
Re: Consecutive integers   [#permalink] 27 Oct 2008, 11:24
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# Consecutive integers

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