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If x is a positive integer, then the least value of x for which x! is

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If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post Updated on: 20 Jan 2019, 06:32
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If x is a positive integer, then the least value of x for which x! is divisible by 1000 is?

A. 5
B. 9
C, 12
D. 15
E. 30

Originally posted by iNumbv on 08 Jan 2015, 08:20.
Last edited by Bunuel on 20 Jan 2019, 06:32, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 08 Jan 2015, 08:32
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iNumbv wrote:
If x is a positive integer, then the least value of x for which x! is divisible by 1000 is?

A. 5
B. 9
C, 12
D. 15
E. 30

Can someone please explain intuitively what the question is asking?


In order x! to be divisible by 1,000, it should have at least 3 trailing zeros. A trailing 0 in factorial of a number is produced by 2 and 5 in it: 2*5 = 10. So, we need 10 to be in x! at least in power of 3.

5! = 120 has 1 trailing zeros.
10! will have 2 trailing zeros.
15! will have 3 trailing zeros.

Answer: D.
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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 08 Jan 2015, 08:28
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Going by logic, x! must contain 1000 or its factors in it

1000= 5*2*5*2*5*2

Always in such type of questions '5' will be the limiting number. So we need three "5" in x!

And smallest factorial will be 15!, it contains 5, 10, 15 .

:-D


iNumbv wrote:
If x is a positive integer, then the least value of x for which x! is divisible by 1000 is?

A. 5
B. 9
C, 12
D. 15
E. 30

Can someone please explain intuitively what the question is asking?

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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 08 Jan 2015, 08:32
iNumbv wrote:
If x is a positive integer, then the least value of x for which x! is divisible by 1000 is?

A. 5
B. 9
C, 12
D. 15
E. 30

Can someone please explain intuitively what the question is asking?



1000= 2^3*5^3

therefore x! must have 5^3 in it. so, out of the given options only 15 and 30 remains. Also, since the question talks about the least value of x. therefore x must be 15.
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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 08 Jan 2015, 08:33
Bunuel wrote:
iNumbv wrote:
If x is a positive integer, then the least value of x for which x! is divisible by 1000 is?

A. 5
B. 9
C, 12
D. 15
E. 30

Can someone please explain intuitively what the question is asking?


In order x! to be divisible by 1,000, it should have at least 3 trailing zeros. A trailing 0 in factorial of a number is produced by 2 and 5 in it: 2*5 = 10. So, we need 10 to be in x! at least in power of 3.

5! = 120 has 1 trailing zeros.
10! will have 2 trailing zeros.
15! will have 3 trailing zeros.

Answer: D.


For similiar questions check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.



Hope this helps.
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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 05 Apr 2018, 08:49
X! will be divisible by 1000 only if it has 3 trailing zeros. 1 trailing zero means there is at least one 2 and one 5 in the product. Concept - https://gmatclub.com/forum/everything-a ... 85592.html

Now, 5 has one trailing zero, 10 has two trailing zeros and 15 has three trailing zeros. So the answer is D.
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Re: If x is a positive integer, then the least value of x for which x! is  [#permalink]

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New post 02 Sep 2019, 22:48
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Re: If x is a positive integer, then the least value of x for which x! is   [#permalink] 02 Sep 2019, 22:48
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