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Let P stand for the product of the first 5 positive integers
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25 Mar 2019, 07:59
25 Mar 2019, 07:59
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Question Stats:
99% (00:48) correct
0% (00:00) wrong based on 14 sessions
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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 \(\frac{P}{10^m}\) is an integer?
(A) 1
(B) 2
(C) 3
(D) 5
(E) 10
(A) 1
(B) 2
(C) 3
(D) 5
(E) 10
Re: Let P stand for the product of the first 5 positive integers
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11 Apr 2020, 11:04
11 Apr 2020, 11:04
Expert Reply
P = 5! (5*4*3*2*1) = 120
So 120/10^M and what is the greatest possible number for M that keeps that fraction an integer?
Try M is 2, which means that the fraction will be 120/100 which is not an integer
Try: M is 1, which means that the fraction will be 120/10 which is 12 an integer
Therefore, Answer is A
So 120/10^M and what is the greatest possible number for M that keeps that fraction an integer?
Try M is 2, which means that the fraction will be 120/100 which is not an integer
Try: M is 1, which means that the fraction will be 120/10 which is 12 an integer
Therefore, Answer is A
gmatclubot
Re: Let P stand for the product of the first 5 positive integers [#permalink]
11 Apr 2020, 11:04
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