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# The product of n consecutive positive integers is always divisible by

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Joined: 01 Nov 2015
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The product of n consecutive positive integers is always divisible by  [#permalink]

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Updated on: 11 Jun 2016, 00:31
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Difficulty:

5% (low)

Question Stats:

93% (01:05) correct 7% (01:39) wrong based on 68 sessions

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The product of n consecutive positive integers is always divisible by

A) n^2 -1
B) (n+1)!
C) 2n +1
D) n^2 + 1
E) n!

Originally posted by aayushagrawal on 10 Jun 2016, 20:27.
Last edited by Bunuel on 11 Jun 2016, 00:31, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The product of n consecutive positive integers is always divisible by  [#permalink]

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11 Jun 2016, 01:05
aayushagrawal wrote:
The product of n consecutive positive integers is always divisible by

A) n^2 -1
B) (n+1)!
C) 2n +1
D) n^2 + 1
E) n!

Plug in some vlues and check -

Product of first 2 number is 2*1

Product of first 3 number is 3*2*1

Product of first 4 number is 4*3*2*1

Product of first 5 number is 5*4*3*2*1

So, The product of first n natural numbers is always divisible by n! , answer will be (E)
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Re: The product of n consecutive positive integers is always divisible by  [#permalink]

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05 Jan 2019, 16:07
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The product of n consecutive positive integers is always divisible by   [#permalink] 05 Jan 2019, 16:07
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