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If n is a positive integer and the product of all integers from 1 to n

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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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New post 20 Oct 2015, 07:17
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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.


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.
So, the answer is D

Cheers,
Brent
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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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New post 03 May 2016, 11:47
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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?

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 from 1 to n  [#permalink]

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New post 14 Aug 2016, 07:04
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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 from 1 to n  [#permalink]

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New post 15 Jan 2018, 13:26
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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 from 1 to n  [#permalink]

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New post 05 Aug 2019, 18:47
Still not sure why is the answer is not 10, since it asks for the "LEAST" possible value of N? Could we say that question is asking for the biggest prime = least possible number?
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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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New post 05 Aug 2019, 19:46
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Hi hedgehuglove,

One of the 'restrictions' in the prompt is that the product has to be a MULTIPLE OF 990. If you multiply the integers from 1 to 10, inclusive, then you will NOT get a multiple of 990. We need to include the number 11 in that group to end up with that multiple.

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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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New post 15 Oct 2019, 19:59
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



We are given that n is a positive integer and the product of all integers from 1 to n, inclusive, is a multiple of 990. Thus:

n! / 990 = integer

To determine the minimum value of n, we need to break 990 into its prime factors.

990 = 10 x 99 = 5 x 2 x 3^2 x 11

Thus:

n! / (5 x 2 x 3^2 x 11) = integer

Since n! must be divisible by 2, 5, 9, and 11, the minimum value of n must be 11. Recall that 11! is divisible by 11 and by any positive integer less than 11. In other words, 11! / 990 = integer.

Answer: B
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Re: If n is a positive integer and the product of all integers from 1 to n   [#permalink] 15 Oct 2019, 19:59

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