All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Factorial zeros (Posted on 2011-02-17)
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.)
 Where n! has the largest proportion of trailing 0s | Comment 1 of 4
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 2011-02-17 19:03:02

 Search: Search body:
Forums (0)