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Let P stand for the product of the first 5 positive integers. What is

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Let P stand for the product of the first 5 positive integers. What is [#permalink]

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New post 17 May 2017, 05:49
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Let P stand for the product of the first 5 positive integers. What is the greatest possible value of m if P/10^m is an integer?

(A) 1
(B) 2
(C) 3
(D) 5
(E) 10
[Reveal] Spoiler: OA

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Last edited by Bunuel on 17 May 2017, 07:15, edited 1 time in total.
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Re: Let P stand for the product of the first 5 positive integers. What is [#permalink]

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New post 17 May 2017, 06:28
Product of first 5 integers = 120
So m should be 1


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Re: Let P stand for the product of the first 5 positive integers. What is [#permalink]

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New post 17 May 2017, 07:58
rohan2345 wrote:
Let P stand for the product of the first 5 positive integers. What is the greatest possible value of m if P/10^m is an integer?

(A) 1
(B) 2
(C) 3
(D) 5
(E) 10


Product of the first 5 positive integers = P = 5! = 120

Given, \(\frac{P}{10^m}\) must be an integer so -

\(\frac{120}{10^m}\) must be an integer

Since p has only 1 trailing zero the value of m can not exceed 1

Thus, answer must be (A) 1
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Re: Let P stand for the product of the first 5 positive integers. What is   [#permalink] 17 May 2017, 07:58
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Let P stand for the product of the first 5 positive integers. What is

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