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.

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

 

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