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        riddles >> medium >> Zeroes of n!
        
(Message started by: ThudanBlunder on Aug 27^th, 2010, 8:26am)

Title: Zeroes of n!
Post by ThudanBlunder on Aug 27^th, 2010, 8:26am

Nay, not with how many naughts doth n! endeth.

Rather, as n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif what fraction of the digits of n! are these end zeroes?

Title: Re: Zeroes of n!
Post by towr on Aug 27^th, 2010, 2:24pm

[hide]There are about n/4 zeroes,
n! is about sqrt((2n+1/3)pi) nⁿ e^-n
So, divide the two and take the limit and you're left with nothing.

Another way to put it, the number of trailing zeroes grow linearly, the number of digits grows exponentially, hence the fraction of trailing zeroes goes to zero.[/hide]

Title: Re: Zeroes of n!
Post by ThudanBlunder on Aug 27^th, 2010, 4:13pm

on 08/27/10 at 14:24:53, towr wrote:

[hide]
n! is about sqrt((2n+1/3)pi) nⁿ e^-n
So, divide the two...[/hide]

No harm in taking logs first...

Title: Re: Zeroes of n!
Post by ThudanBlunder on Aug 29^th, 2010, 1:20pm

on 08/27/10 at 14:24:53, towr wrote:

[hide]
...the number of digits grows exponentially....[/hide]

NOT...

Title: Re: Zeroes of n!
Post by towr on Aug 29^th, 2010, 1:38pm

In order of the size of the input they do.

[edit]Well, okay, it doesn't really help to use different measures in both terms..

So, rather let's just take Stirling's approximation, and go with [n/4] / [n ln(n)-n] -> 0 as n -> inf.