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

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When k! is written as m*10^n so that the last digit of [#permalink] New post 04 Oct 2006, 06:51
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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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 [#permalink] New post 05 Oct 2006, 01:43
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
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Last edited by cicerone on 25 Sep 2008, 00:32, edited 1 time in total.
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 [#permalink] New post 05 Oct 2006, 07:12
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?
  [#permalink] 05 Oct 2006, 07:12
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When k! is written as m*10^n so that the last digit of

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