At 09:30 PM Sunday 4/17/2005, Julia Thompson wrote:
Ronn!Blankenship wrote:
At 08:49 PM Sunday 4/17/2005, Alberto Monteiro wrote:
Ronn!Blankenship wrote: > >> Yes, 2500! = 1.628 10^7411 has 7412 digits, but 500 of >> them are zeroes > > 624, actually. > Ah, ok, I should have mentioned that 500 of them are _trailing_ zeroes. I didn't count the middle zeroes
Neither did I . . .
2500! has 500 trailing 0s from the 500 numbers divisible by 5 another 100 trailing 0s from the 100 numbers divisible by 25 another 20 trailing 0s from the 20 numbers divisible by 125 another 4 trailing 0s from the 4 numbers divisible by 625 for a total of 624 trailing 0s.
(Of course, you need a 2 to go with each 5 to give you a trailing 0, but as 2500! has as a factor 2^2495 (if I did my math right), that shouldn't be too much of a problem....)
Heck, I just counted 'em . . .
Actually I Told The Computer To Count Them For Me Maru
Telling the computer to count them is probably the most efficient method. :)
I just felt like explaining. I'm waiting for Alberto's response now.
Julia _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
