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If n is a positive integer such that n! is divisible by 840, what is

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Bunuel
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If n is a positive integer such that n! is divisible by 840, what is [#permalink] New post   20 Mar 2018, 01:00
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If n is a positive integer such that n! is divisible by 840, what is the least possible value of n?

(A) 7
(B) 8
(C) 12
(D) 14
(E) 15
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A
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DavidTutorexamPAL
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Re: If n is a positive integer such that n! is divisible by 840, what is [#permalink] New post   20 Mar 2018, 01:43
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Bunuel wrote:
If n is a positive integer such that n! is divisible by 840, what is the least possible value of n?

(A) 7
(B) 8
(C) 12
(D) 14
(E) 15


To answer this we'll first need to factorize 840.
This is a Precise approach.

840 = 210*4 = 7*3*10*4 = 7*3*5*2*4. Rearranging, we have 7*5*4*3*2, so 7!, is enough.

(A) is our answer.
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If n is a positive integer such that n! is divisible by 840, what is [#permalink] New post   20 Mar 2018, 10:36
Bunuel wrote:
If n is a positive integer such that n! is divisible by 840, what is the least possible value of n?

(A) 7
(B) 8
(C) 12
(D) 14
(E) 15


\(840 = (2^3)*3*5*7\).

We got n! is divisible by 840, therefore the least number of n is 7 (7! got 3-2's from 2,4,6; 1-3's from 3 or 6; 1-5's and 1-7's)
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Re: If n is a positive integer such that n! is divisible by 840, what is [#permalink] New post   05 Apr 2018, 11:07
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Bunuel wrote:
If n is a positive integer such that n! is divisible by 840, what is the least possible value of n?

(A) 7
(B) 8
(C) 12
(D) 14
(E) 15


Let’s factor 840:

840 = 84 x 10 = 21 x 4 x 5 x 2 = 7 x 3 x 4 x 5 x 2 . Note that the greatest prime factor of 840 is 7.

Since 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1, we see that all of the factors of 840 are also factors in 7!, so n = 7 is the least possible value of n such that n! is divisible by 840.

Answer: A
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Re: If n is a positive integer such that n! is divisible by 840, what is [#permalink] New post   17 Apr 2020, 11:32
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Re: If n is a positive integer such that n! is divisible by 840, what is [#permalink]
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