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# If n is a positive integer and the product of all integers

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If n is a positive integer and the product of all integers [#permalink]  28 Dec 2009, 10:47
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Question Stats:

82% (01:37) correct 18% (00:57) wrong based on 301 sessions
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:
I got the right answer 11 by stating:
--> 990 can be divided by 11

Is is the right way to solve that?

[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Feb 2014, 03:56, edited 2 times in total.
Edited the question and added the OA
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Re: Math Questions [#permalink]  28 Dec 2009, 11:22
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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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.
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Re: Math Questions [#permalink]  28 Dec 2009, 11:33
Thanks very Bunuel !!!

It makes sense.
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Re: GMAT Prep q#11 need help [#permalink]  27 Feb 2010, 20:50
3
KUDOS
divanshuj wrote:
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 for n?

a) 10
b) 11
c) 12
d) 13
e) 14

990 = 2 * 5 * 3 * 3* 11
so product from 1 to n should have all the above numbers. n = 11 will have all the above numbers as factors of 1 * 2* ..... * 11

B
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Re: n is a positive int and the product of all ints from 1 to n [#permalink]  01 Aug 2010, 23:00
Hi,

The question says that 990 is a factor of n!.
990 = 9*110 = 9*11*10
For 990 to be a factor of n!, n should at least be 11 so that it includes the common factor 11 that is found in 990.

regards,
Jack
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Re: n is a positive int and the product of all ints from 1 to n [#permalink]  02 Aug 2010, 06:18
1
KUDOS
990= 2x3x3x5x11
So 11 has to be one of the prime factor
So 11!
_________________

+1 kudos me if this is of any help...

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Re: Multiple of 990 [#permalink]  18 Sep 2010, 14:18
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This question tests the concept of prime factorization. In order to figure out if 990 can be a divisor of the unknown product you need to determine all of the prime factors that combine to successfully divide 990 without leaving a remainder.

The prime factors of 990 are 2, 3, 3, 5, 11

Since 11 is a prime factor of 990, you need to have the number present somewhere in the unknown product. Answer A is the product of integers from 1 to 10 and therefore does not meet the necessary criteria. Answer B includes 11 and therefore is the answer.
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Re: Multiple of 990 [#permalink]  18 Sep 2010, 14:59
The least possible value of such a n will be at least the highest prime factor of 990, which is 11

The only other thing to check is if the rest of the factors of 990, can be obtained by multiplication of numbers less than 11. Otherwise n may be greater than 11

But that is easy to see since 990=9*10*11. So taking all the numbers below 11 will suffice.
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If n is a positive integer... [#permalink]  13 Feb 2011, 17:43
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?

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

How might one go about solving this question quickly? I tried solving this one and it started taking way too long.
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Re: If n is a positive integer... [#permalink]  13 Feb 2011, 17:53
Expert's post
1
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Merging similar topics.

Also check: property-of-integers-104272.html
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Re: Math Questions [#permalink]  21 Jul 2011, 22:49
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

Why is this k introduced? Whats the concept of introducing k ?? n ! = 990 * k ?? I am confused...
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Re: Math Questions [#permalink]  11 Apr 2012, 01:46
Expert's post
siddhans wrote:
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

Why is this k introduced? Whats the concept of introducing k ?? n ! = 990 * k ?? I am confused...

We are told that n! is a multiple of 990, which can be expressed as n!=990*k, for some positive integer multiple k.
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Re: Math Questions [#permalink]  27 Oct 2012, 11:04
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

out of curiousness, i would like to clarify something here.

In case if we have a option like (i know option are pathetic)

A. 2
B. 5
C. 11
D. 13
E. any option..

then what would be the answer.. BTW when i try to solve this question i was only having the question(without option) , I took the prime factor stuff & chosen the least one from that.. please help me here to understand more.
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Re: Math Questions [#permalink]  29 Oct 2012, 05:25
Expert's post
breakit wrote:
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

out of curiousness, i would like to clarify something here.

In case if we have a option like (i know option are pathetic)

A. 2
B. 5
C. 11
D. 13
E. any option..

then what would be the answer.. BTW when i try to solve this question i was only having the question(without option) , I took the prime factor stuff & chosen the least one from that.. please help me here to understand more.

We are asked about the least possible value of n. The least value of n is 11. Answer choices have nothing to do with that.
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Re: Math Questions [#permalink]  25 Jan 2013, 01:29
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

to cement my understanding, maximum value of n can be anything, right?
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Re: Math Questions [#permalink]  25 Jan 2013, 03:51
1
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Expert's post
Sachin9 wrote:
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

to cement my understanding, maximum value of n can be anything, right?

The maximum value of n is not limited at all. For all n more than or equal to 11, n! will be a multiple of 990.
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Re: Math Questions [#permalink]  14 Jul 2013, 23:29
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

Hi Bunuel,

I am not sure why have excluded 10, which seems to be one of the factors (5*2).
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Re: Math Questions [#permalink]  14 Jul 2013, 23:33
1
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Expert's post
davidfrank wrote:
Bunuel wrote:
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

We are told that $$n!=990*k=2*5*3^2*11*k$$ --> $$n!=2*5*3^2*11*k$$ 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)..

Hope it's clear.

Hi Bunuel,

I am not sure why have excluded 10, which seems to be one of the factors (5*2).

n! must be a multiple of 990=90*11. Now, if n=10, then it wont have 11 as its factor, so it cannot be a multiple of a number which HAS 11 as its factor.

Hope it's clear.
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Re: If n is a positive integer and the product of all integers [#permalink]  19 Jul 2014, 16:50
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Re: If n is a positive integer and the product of all integers [#permalink]  19 Oct 2014, 15:11
can somebody explain this in plain English. I think I can follow what people are saying but I don't really understand the concept so I'm not sure i'll be able to apply it to another problem. I don't want to know why 11 but what is the thought process/logic behind this question (much props to explanations without lengthy formulas).
Re: If n is a positive integer and the product of all integers   [#permalink] 19 Oct 2014, 15:11

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