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If (20!*20!)/20^n is an integer

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If (20!*20!)/20^n is an integer  [#permalink]

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New post 07 Jan 2019, 10:52
1
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

61% (01:19) correct 39% (01:55) wrong based on 53 sessions

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If (20!*20!)/20^n is an integer, what is the largest possible value of n?

A. 6
B. 8
C. 10
D. 11
E. 12


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Posts: 64
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If (20!*20!)/20^n is an integer  [#permalink]

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New post Updated on: 08 Jan 2019, 00:10
2
anshkool4u wrote:
If (20!*20!)/20^n is an integer, what is the largest possible value of n?

A. 6
B. 8
C. 10
D. 11
E. 12


Please Press Kudos if you find the question worth it.


\(\frac{(20!*20!)}{20^n}\)
Greatest possible value of 'n' for which the fraction will be an integer means finding the total number of "20s" in 20! x 20!

20 = 5 x \(2^2\)
Since the number of 2s will always be more than number of 5s, we calculate only the number of 5s in 20!

Number of 5s in 20! = \(\frac{20}{5}\) = 4
Since there are 2 20! in numerator, the number of 5s in 20^n will be 2x4 = 8

Hence option B.

Alternate Method


\(20!*20! = (20!)^2\)
\(\frac{(20!)^2}{20^n}\) is an integer and value of 'n' will be the same as the maximum number of 20s in \((20!)^2\)

Number of 20s means finding number of 5s.
Therefore, 20/5 = 4
So n=4
i.e \(20^4\)
Number of 5s (or 20s) in \((20!)^2 = (20^4)^2 = 20^8\)

i.e n=8. Option B
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Originally posted by Darshi04 on 07 Jan 2019, 11:53.
Last edited by Darshi04 on 08 Jan 2019, 00:10, edited 1 time in total.
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Re: If (20!*20!)/20^n is an integer  [#permalink]

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New post 07 Jan 2019, 13:14
Finding out number of 5's in 20! Will give us 4.
And as 20! Is given twice we take 4+4=8.
Hence, B.

Why we look for 5 and not 2?
→ because there will always be a greater no of 2's so we look for exact number of 5's to know the perfect answer.

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Re: If (20!*20!)/20^n is an integer   [#permalink] 07 Jan 2019, 13:14
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