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If 20! × 20!/20^n is an integer, what is the largest possible value of

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If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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New post 28 Apr 2016, 16:00
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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Bunuel wrote:
If 20! × 20!/20^n is an integer, what is the largest possible value of n?

A. 20
B. 16
C. 8
D. 4
E. 1


20! × 20!/20^n is an integer.
Or 20! × 20!/(2^2n*5^n)

This means we need to find the power of 5 in the numerator. We can safely say that the number of powers of 5 will be lower than the number of powers of 4 in the numerator.

Largest power of 5 in 20! = [20/5] + [20/25] = 4
Since there are 2 20!'s, power of 4 in the numerator = 2*4 = 8

Hence the largest value of b for which 20! × 20!/20^n is integer = 8

Correct Option : C
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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New post 29 Apr 2016, 10:16
in 20! following can be factorized as 20 :
1*20, 2*10, 4*5

also another factor from 5*3*8 can be divided by 20
so n = 8

Ans: C
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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New post 04 Nov 2016, 07:41
This is an Excellent Question.
Here number of 5's plays a limiting role.
As the number of 4's would be sufficient. (ample number of 2's)
number of 5's in 20! => 20/5+20/25 => 4
therefore 20!*20! would have 8 fives and ample number of 4's
so since a 20 is formed by a 5 and a +> Maximum number of 20's => 8
N can be almost 8

Hence C
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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New post 14 Jan 2018, 15:10
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stonecold wrote:
This is an Excellent Question.
Here number of 5's plays a limiting role.
As the number of 4's would be sufficient. (ample number of 2's)
number of 5's in 20! => 20/5+20/25 => 4
therefore 20!*20! would have 8 fives and ample number of 4's
so since a 20 is formed by a 5 and a +> Maximum number of 20's => 8
N can be almost 8

Hence C


could you explain how 20/5 +20/25 equals four?
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink]

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New post 14 Jan 2018, 19:32
destinyawaits wrote:
stonecold wrote:
This is an Excellent Question.
Here number of 5's plays a limiting role.
As the number of 4's would be sufficient. (ample number of 2's)
number of 5's in 20! => 20/5+20/25 => 4
therefore 20!*20! would have 8 fives and ample number of 4's
so since a 20 is formed by a 5 and a +> Maximum number of 20's => 8
N can be almost 8

Hence C


could you explain how 20/5 +20/25 equals four?


You should take only the quotient of the division, that is 20/5 = 4 and 20/25 = 0.
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of   [#permalink] 14 Jan 2018, 19:32
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