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# When k! is written as m*10^n so that the last digit of

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Joined: 04 Jul 2006
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When k! is written as m*10^n so that the last digit of [#permalink]

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04 Oct 2006, 07:51
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When k! is written as m*10^n so that the last digit of integer m is not 0, the difference between the number of digits in m and the value of n is 5. If t is the number of possibilities there are the value of the integer k, then t+1=

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

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Senior Manager
Joined: 28 Aug 2006
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05 Oct 2006, 02:43
Is it B.

12! is the only number which satisfies this condition.

ie 12!= mx10^2 and m contains 7 digits. so m-n ie 7-2 =5
_________________

Last edited by cicerone on 25 Sep 2008, 01:32, edited 1 time in total.

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Intern
Joined: 21 Sep 2006
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05 Oct 2006, 08:12
cicerone wrote:
Is it B.

12! is the only number which satisfies this condition.

ie 12!= mx10^2 and m contains 7 digits. so m-n ie 7-2 =5

Cicerone - How did you get to 12! so quickly?

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05 Oct 2006, 08:12
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