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Consecutive integers

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Consecutive integers [#permalink] New post 24 Oct 2008, 14:42
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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Re: Consecutive integers [#permalink] New post 24 Oct 2008, 20:43
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 :)
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Re: Consecutive integers [#permalink] New post 25 Oct 2008, 00:02
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.
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Re: Consecutive integers [#permalink] New post 25 Oct 2008, 00:10
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 ??
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Re: Consecutive integers [#permalink] New post 25 Oct 2008, 00:53
Amit u r right. SImple calculation error. I couldn;t solve 32-6 =? equation. :oops: 26! it is .
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Re: Consecutive integers [#permalink] New post 27 Oct 2008, 09:10
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.
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Re: Consecutive integers [#permalink] New post 27 Oct 2008, 10:24
TYpo. sorry. u indeed not mentioned 9-my bad.
Re: Consecutive integers   [#permalink] 27 Oct 2008, 10:24
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