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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: ( Back to comment list | You must be logged in to post comments.)
Two answers: | Comment 3 of 4 |

Two answers: 5! and 6!  have 33.3 % ratio.

 


  Posted by Ady TZIDON on 2011-02-18 04:36:32
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