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.
There is a trailing zero in
n! for each factor of 5 in
n.
The rate at which number of trailing zeroes increases is at a slower rate than the rate at which the length of the number increases. Thus, the greatest proportion of trailing zeroes to the length of
n is where
n=5. At this point, the length of
n is 3 and the number of trailing zeroes is 1, giving a 33 1/3% proportion of trailing zeroes to the length of
n.

Posted by Dej Mar
on 20110217 19:03:02 