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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)..

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.
_________________

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.

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.
_________________

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).

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.

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).

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

Answer: option D
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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.

Answer D.
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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)

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. So, the answer is B

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

Is is the right way to solve that?

Thanks in advance.

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.

Answer: B
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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

This post received KUDOS

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.