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

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Math Expert
Joined: 02 Sep 2009
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29 Oct 2012, 06:25
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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25 Jan 2013, 02:29
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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25 Jan 2013, 04:51
1
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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15 Jul 2013, 00:29
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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15 Jul 2013, 00:33
2
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]

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19 Oct 2014, 16:11
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).
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If n is a positive integer and the product of all the integers from 1  [#permalink]

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20 Oct 2015, 04:31
1
1
Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

Kudos for a correct solution.

Bunuel: I guess, I have seen this question on GMAT CLUB before
if-n-is-a-positive-integer-and-the-product-of-all-integers-90855.html

product of all the integers from 1 to n = n!
990 = 9*10*11
i.e. n! must be divisible by 9, 10 and 11 all for n! to be divisible by 990
i.e. minimum value of n must be 11 for n! to be divisible by 11 and 10 and 9 both

P.S. 11!= 11*10*9*8*7*6*5*4*3*2*1

n = 11

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Re: If n is a positive integer and the product of all the integers from 1  [#permalink]

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20 Oct 2015, 05:28
1
Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

Kudos for a correct solution.

My Solution:

Prime factorization of 990 gives us = 2 x 3 x 3 x 5 x 11

Therefore, if n! is divisible by 990 it must contain all above factors.

Option A , B and C don't contain 11 as factor. So least value of n must be 11 as 11! contains all possible factors.

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Re: If n is a positive integer and the product of all the integers from 1  [#permalink]

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20 Oct 2015, 06:05
Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

Kudos for a correct solution.

First of all, select all prime factors of 990. Those are 2, 5, 3^2 and 11. Since the product of 1 to n is divisible by 990, it has all prime factors of 990. Therefore the least possible value of n is 11.

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Re: If n is a positive integer and the product of all the integers from 1  [#permalink]

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20 Oct 2015, 07:17
1
Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

Kudos for a correct solution.

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)

For more on this concept, see the 2nd half of this free video: http://www.gmatprepnow.com/module/gmat- ... /video/825

So, if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.

Cheers,
Brent
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Re: If n is a positive integer and the product of all the integers from 1  [#permalink]

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20 Oct 2015, 22:40
Bunuel wrote:
If n is a positive integer and the product of all the integers from 1 to n, inclusive, is divisible by 990, what is the least possible value of n?

(A) 8
(B) 9
(C) 10
(D) 11
(E) 12

Kudos for a correct solution.

990 = 2 * 3 * 3 * 5 * 11

so integers less than n should contain all factors 990

there fore least value of n should be 11

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If n is a positive integer and the product of all the integers from 1  [#permalink]

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23 Oct 2015, 07:51
990= 3^2 * 11 * 2 * 5

Product of all the integers from 1 to n = factorial (n). Since n! is divisible by 990, its expansion should contain all factors of 990.

Since we need the least possible value of n, the answer is 11.

Option D.
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Re: If n is a positive integer and the product of all integers  [#permalink]

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09 Jan 2016, 10:54
Quote:
If n is a positive integer and 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

A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N
Similarly, we can say:
If N is is a multiple of k, then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = (2)(2)(2)(3)
70 is divisible by 5 <--> 70 = (2)(5)(7)
330 is divisible by 6 <--> 330 = (2)(3)(5)(11)
56 is divisible by 8 <--> 56 = (2)(2)(2)(7)

So, if if some number is a multiple of 990, then 990 is hiding in the prime factorization of that number.

Since 990 = (2)(3)(3)(5)(11), we know that one 2, two 3s, one 5 and one 11 must be hiding in the prime factorization of our number.

For 11 to appear in the product of all the integers from 1 to n, n must equal 11 or more.

Cheers,
Brent
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Re: If n is a positive integer and the product of all integers  [#permalink]

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03 May 2016, 11:47
1
nailgmattoefl 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

I got the right answer 11 by stating:
--> 990 can be divided by 11

Is is the right way to solve that?

Solution:

We are given that the product of all integers from 1 to n, inclusive, (i.e., n!) is a multiple of 990.

Thus, n! = 990 x k for some integer k.

Another way to visualize the given information is to say:

n!/990 must be an integer. Thus, n! must contain all the factors of 990.

Notice that 990 = 99 x 10 = 9 x 11 x 10. So in order for n!/990 to be an integer, n must at least be equal to 11, since 11! = 11 x 10 x 9 x … x 3 x 2 x 1.

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Re: If n is a positive integer and the product of all integers  [#permalink]

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14 Aug 2016, 07:04
1
1
Lets breakdown the question:
n is a positive integer
Therefore n>0

the product of all the integers from 1 to n, inclusive, is a multiple of 990

Therefore (1 x 2 x 3 ..... x n) / 990 gives an integer
Therefore n!/990 gives an integer
Therefore n!/(9 x 10 x 11)gives an integer

Question: What is the least possible value of n!
This means what is the lowest value that n! can be so that when it is divided by 9 x 10 x 11, the outcome gives an integer

The lowest value of n! should have all the factors that are in the demoninator so that when divided an integer is formed

11! / (9 x 10 x 11) = (11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (9 x 10 x 11)
= 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

10! / (9 x 10 x 11) = (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (9 x 10 x 11)
= (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / 11 = mixed number

Any number lower than 10! will give a mixed number.

Any number higher that 11! will give an integer, however 11! is the lowest value that will produce an integer when 990 is the denominator.
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Re: If n is a positive integer and the product of all integers  [#permalink]

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26 Sep 2017, 13:10
Question can be restated as for what smallest value of n is n! a factor of 990

990 = 3.3.11.2.5

minimum we need 11! for 990 hence the answer is 11.
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Re: If n is a positive integer and the product of all integers  [#permalink]

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26 Nov 2017, 04:15
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.

Bunuel hello! is there a shortcut for prime factorization ? for example you wrote $$n!=990*k=2*5*3^2*11*k$$ - Did you write it after using prime factorization tree method ? Or some other method ? thanks and have a great day! D
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Re: If n is a positive integer and the product of all integers  [#permalink]

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04 Jan 2018, 15:31
Why is it not 3^2=9?

Posted from my mobile device
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Re: If n is a positive integer and the product of all integers  [#permalink]

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04 Jan 2018, 19:06
SamDGold wrote:
Why is it not 3^2=9?

Posted from my mobile device

---------------------

3^2 is not equal to 9, because 990 was broken into its prime factors. (9 is not a prime number, hence expressed as 3^2)
eg. 48 =(2^4)*(3^1)
here 2,3 are prime factors and 48 expressed in terms of 2 & 3 and not as 2^4 which would be simple factorization

As for solution:
The question asks for product of all integers from 1 to n (inclusive) => which is n!

Now to reframe the question: For which value of does n! be a multiple 990 ?
On looking at 990, we can break it as
990 =99 * 100
=(9*11)*10
=9*10*11
so you need to have 11,10 & 9 appearing in n!, which is only possible for n>=11
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Re: If n is a positive integer and the product of all integers  [#permalink]

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15 Jan 2018, 13:26
Hi All,

For a number to be a multiple of 990, that number must have the exact same prime factors (including duplicates) as 990 (but may have "extra" prime factors as well).

This prompt adds the extra stipulation that N has to be as SMALL as possible, so after factoring 990, we need to find the smallest product that "holds" the 2, 5, 11, and two 3s that make up 990.

1(2)(3)(4)....(11) is the smallest product that does that, so N = 11.

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Re: If n is a positive integer and the product of all integers   [#permalink] 15 Jan 2018, 13:26

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