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Intern  Joined: 18 Dec 2009
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If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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Question Stats: 80% (01:14) correct 20% (01:24) wrong based on 2981 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

Spoiler: :: This question has different answer choices in OG 2020
A. 8
B. 9
C. 10
D. 11
E. 12

Originally posted by nailgmattoefl on 28 Dec 2009, 10:47.
Last edited by Bunuel on 07 Jul 2019, 22:23, edited 4 times in total.
Edited the question and added the OA
Math Expert V
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Posts: 65785
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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1
59
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: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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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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Manager  Joined: 26 May 2005
Posts: 141
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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6
2
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: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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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
Intern  Joined: 06 Oct 2009
Posts: 43
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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3
990= 2x3x3x5x11
So 11 has to be one of the prime factor
So 11!
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Location: London
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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

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3
Similar question: https://gmatclub.com/forum/n-is-a-posit ... 04272.html
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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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5
You need to primarily find the primes of 990:
2,3,3,5,11
The integer must contain all these primes

By knowing 11 is an prime, we know the answer can be B,C,D or E as these all contain 11.
11=1,2,3,4,5,6,7,8,9,10,11

11 contains the 2,3,5 and 11 explicitly and the second 3 comes from the 6 (2*3), therefore the answer is 11.

This is because we know a number that contains all the primes of 990 will be a multiple of 990.

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

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1
The answer would be the number which contains all the factors of the figure.
The series of 1 to 11 contained all the factors of 990. 2,3,3,5 and 11.
Because 990 is made up from these 5 numbers such that 2*3*3*5*11 = 990
If you times the other numbers (1,4,6,7,8,9,10) in the series you will get a multiple of 990 from this.

I picked 11 because it asked for the least possible value of n. Picking 12,13,14 would result in a multiple of 990 but the number is more than the least possible value of 11.

The best way to interpret problems is through practice, you will come to understand questions better over time. Always remember GMAT questions are simple questions made to look complicated.
Intern  Joined: 09 Feb 2011
Posts: 8
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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1
just entered this in google and you're right! (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11) / 990 = 40 320

meanwhile (1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10) / 990 = 3 665.45455 wouldn't work

It makes sense because the highest prime factor of 990 must also be a factor of any multiple of 990.

I guess I need to do more practice problems involving multiples and number properties.
Manager  Joined: 20 Nov 2010
Posts: 113
Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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Since we need to find the minimum value of n, So the greatest prime factor of 990 will be the answer.
990 = 2 * 3^2 * 5 * 11.
Hence the minimum value of n such that n! ( 1*2* ... *n) is divisible by 990 is 11.
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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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

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

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

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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 the  [#permalink]

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Here's a more elaborated thinking response to what you all have summarized above -- but for those who need the extra steps, try this.

990* multiple = product of 1 to n
990 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n
11 * 3 * 3 * multiple = 1 * 2 * 3 * 4 * 5 * ... * n

We need to find the smallest value for n. We know in the factor tree, any number can be broken down to its lowest level in the factor tree. And that factor tree will comprise of all prime factors.

The factors on the left side are:
11 and 3 and 'multiple' -- the multiple can be anything.

We know that we want 990 to go into the right side (we don't care how many times).

The 990 consists of 11 and 3^2. So essentially it's prime factors are 11 and 3. These lowest level prime factors must go into the right side -- so the right side at the very least must contain the lowest prime factors from the left side.

So, we see 1 * 2 * 3 * 4 * 5 * ... * n

contains a 3.

But it does not contain an '11'.

Hope that helps.
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Re: If n is a positive integer and the product of all the  [#permalink]

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flood wrote:
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

I got this one right but I'm not too clear on the logic behind it.

Can you multiply two numbers (other than 11) and get an 11? No because 11 is prime. If 11 is a part of the product of 3 numbers, say a*b*c, it must have been one of the numbers.
Given that the product is 990x (where x is some integer). When we start breaking down 990 into factors, we start with 9*11*10. We see that 9 and 10 will be further broken down into smaller factors but 11 will not be. It is prime. So one of the numbers which multiplied to give 990x must have been 11.
Since consecutive integers 1 to n were multiplied, n is either 11 or greater than 11 to include 11 in the numbers multiplied.
Least possible value of n is 11.
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Re: If n is a positive integer and the product of all integers from 1 to n  [#permalink]

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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.

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 integers from 1 to n  [#permalink]

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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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"Do not watch clock; Do what it does. KEEP GOING." Re: If n is a positive integer and the product of all integers from 1 to n   [#permalink] 20 Oct 2015, 04:28

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