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

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Bunuel
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If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  28 Apr 2016, 17:00
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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
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OptimusPrepJanielle
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  29 Apr 2016, 00:02
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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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nipunjain14
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  29 Apr 2016, 11:16
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] New post  04 Nov 2016, 08:41
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] New post  14 Jan 2018, 16:10
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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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Bunuel
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  14 Jan 2018, 20:32
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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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If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  Updated on: 16 Sep 2020, 18:52
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As 20=2^2 * 5
(20!)^2 = (1*2*...*20)^2 = (2^10*5^4*...)^2 = (20^4*...)^2 = 20^8*...
Hence C.

Originally posted by dieplengoc on 16 Sep 2020, 00:23.
Last edited by dieplengoc on 16 Sep 2020, 18:52, edited 2 times in total.
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  16 Sep 2020, 03:22
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\(20^n\) = \((2 * 2 * 5)^n\)

=> 20 = \((2^2)^n * 5^n\)

So, we need a number of 5's in 20! to find the value of n.

=> \(\frac{20}{5}\) + \(\frac{(20) }{ 5^2 }\)

=> 4 + 0

=> So there will be total 4 + 4 = 8 (5's) for [20! * 20!]

Therefore n = 8

Answer C
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datbanhbao
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If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  Updated on: 18 Aug 2021, 03:42
I dont know this will this solution be correct or not? But I hope I can share to you because I got wrong with this quiz

20! x 20!/20^n = Integer >> this one has to be divisible >> we have: (20!)^2/20^n = integer
20! will have 4 pairs that can divide 20: 1-20; 2-10; 4-5; 15-8 >> (20!)^2 will have 8 pairs >> n = 8 >> C

Originally posted by datbanhbao on 04 Aug 2021, 09:28.
Last edited by datbanhbao on 18 Aug 2021, 03:42, edited 1 time in total.
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  11 Aug 2021, 10:29
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! will have 4 20s
since we can only make 20s
with 5 and only 4 multiples 5, 10 ,15,20 exists
THey can be made 20 with the it's respective multiple of 2
and 2 20! exists therefore the total number becomes 8
Therefore IMO C
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AJ0784
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Re: If 20! × 20!/20^n is an integer, what is the largest possible value of [#permalink] New post  14 Aug 2021, 12:36
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


Think in this way to understand this question:
For 625/5^n to be an integer the max value of n can be 4 since 625=5^4
Similarly for 20!*20!/20^4 to be an integer we have to find out the max value of the power of 5 in 20!*20!.
We have 4 fives in 20! Therefore 20!*20!has 8 fives.Hence the maximum value of n can be 8.

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