If n is a positive integer and the product of all integers : GMAT Problem Solving (PS)
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# If n is a positive integer and the product of all integers

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Intern
Joined: 08 Mar 2009
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If n is a positive integer and the product of all integers [#permalink]

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07 Nov 2009, 10:02
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25% (medium)

Question Stats:

83% (01:29) correct 17% (00:52) wrong based on 18 sessions

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If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10
B. 11
C. 12
D. 13
E. 14
[Reveal] Spoiler: OA

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Math Expert
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Kudos [?]: 93070 [0], given: 10541

Re: If n is a positive number Gmat prep [#permalink]

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07 Nov 2009, 12:18
If n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?
A. 10
B. 11
C. 12
D. 13
E. 14

$$990=2*5*3^2*11$$

We are told that $$n!=990*k=2*5*3^2*11*k$$ which means that n! must have all factors of 990 to be the multiple of 990, hence must have 11 too, so the least value of n is 11 (notice that 11! will have all other factors of 990 as well, otherwise the least value of n would have been larger).

Answer: B.
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If n is a positive [#permalink]

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12 Dec 2011, 05:47
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is a multiple of 990, what is the least possible value of n?

Guys - any idea how to solve this questions? OA is not provided unfortunately.
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Re: Multiple of 990 [#permalink]

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12 Dec 2011, 06:55
is it 11 ?
990 = 2 * 5 * 9 * 11
so the least value for "N" can be 11.
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Re: Multiple of 990 [#permalink]

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12 Dec 2011, 22:12
990= 11*10*9

since 11 is a prime no which does not have factor apart from 11 and 1. so 11 is necessary no for 990.

hence n=11
Re: Multiple of 990   [#permalink] 12 Dec 2011, 22:12
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# If n is a positive integer and the product of all integers

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