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Re: Math Questions [#permalink]
28 Dec 2009, 11:22

3

This post received KUDOS

Expert's post

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)..

Re: GMAT Prep q#11 need help [#permalink]
27 Feb 2010, 20:50

3

This post received 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

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.

Re: Multiple of 990 [#permalink]
18 Sep 2010, 14:18

2

This post received KUDOS

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.

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.
_________________

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)..

Answer: B.

Hope it's clear.

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

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)..

Answer: B.

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.
_________________

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)..

Answer: B.

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.

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)..

Answer: B.

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.
_________________

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)..

Answer: B.

Hope it's clear.

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

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: Math Questions [#permalink]
25 Jan 2013, 03:51

1

This post received KUDOS

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)..

Answer: B.

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.
_________________

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)..

Answer: B.

Hope it's clear.

Hi Bunuel,

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

Re: Math Questions [#permalink]
14 Jul 2013, 23:33

1

This post received KUDOS

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)..

Answer: B.

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.