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Factorial zeros (Posted on 2011-02-17) Difficulty: 2 of 5
41! = 33452526613163807108170062053440751665152000000000
which is 50 digits long, the last 9 of them are 0s. Thus the trailing zeros make up 18% of the entire number.

Find n where n! has the largest possible proportion of trailing 0s.

Prove it.

No Solution Yet Submitted by Jer    
Rating: 5.0000 (1 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Possible solution(spoiler)Danish Ahmed Khan2012-10-27 12:11:50
Two answers:Ady TZIDON2011-02-18 04:36:32
SolutionExploration and an attempt at proof.Charlie2011-02-17 20:39:20
Where n! has the largest proportion of trailing 0sDej Mar2011-02-17 19:03:02
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