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If n! = n * (n-1) * (n-2)... 1, where n is a positive integer

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If n! = n * (n-1) * (n-2)... 1, where n is a positive integer  [#permalink]

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New post 28 Aug 2018, 13:26
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Question Stats:

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If n! = n * (n-1) * (n-2)... 1, where n is a positive integer, which of the following is (are) true?

I 26 is not a factor of 25!
II 29 is a factor of 25!
III 1000000 is a factor of 25!

A. None
B. III Only
C. I and II only
D. II and III only
E. I, II, and III

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Re: If n! = n * (n-1) * (n-2)... 1, where n is a positive integer  [#permalink]

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New post 28 Aug 2018, 18:13
n! = n * (n-1) * (n-2)... 1, where n is a positive integer, which of the following is (are) true?

Since it asks for correctness of three cases, let's take these cases one by one..

I 26 is not a factor of 25!
25! Is multiple of all numbers less than and equal to 25.
Also 26=2*13 but 25!=1*2*3*.....*12*13*14*.....24*25
25! Is a multiple of both 2 and 13 thus multiplof 2*13 or 26.
Hence the statement is not true.

II 29 is a factor of 25!
29 is a prime number greater than 25 ..
But 25 is a multiple of only prime numbers till 25. Hence 29 is not a factor.
Not true.

III 1000000 is a factor of 25!
1000000=\(10^6=2^65^6\)
So let's check if there are atleast six 5s in 25!
# of 5s =\(\frac{25}{5}+\frac{25}{25}=5+1=6\)
Therefore 25! Contains six 5s and thus 1000000 is a factor of 25!
True

Only III true


A. None
B. III Only
C. I and II only
D. II and III only
E. I, II, and III


B
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Re: If n! = n * (n-1) * (n-2)... 1, where n is a positive integer  [#permalink]

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New post 28 Aug 2018, 23:31
1. 26 is a factor of 25!
26= 13*2 and in 25! we have both 13 and 22
2. 29 is a prime number and is greater than 25. Hence it cannot be the factor of 25!
3. No of trailing zeros in 25!
25!/5 = 5
5/5 =1
Hence total no of zeroes in 25! = 6 (we donot need to calculate the power of 2 as it is sufficient to know the power of 5 (logic : 5*2 =10))
Hence 1000000 is a factor of 25!
The answer is B
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Re: If n! = n * (n-1) * (n-2)... 1, where n is a positive integer   [#permalink] 28 Aug 2018, 23:31
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