Find the last zeros of $2011!$

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Moon
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Find the last zeros of $2011!$

Unread post by Moon » Thu Jan 06, 2011 7:42 pm

In how many zeros does $2011!$ end?

BTW is there any junior people here? I am curious. :)
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Labib
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Re: Find the last zeros of $2011!$

Unread post by Labib » Thu Jan 06, 2011 10:46 pm

I'm not Junior, but I like this prob: so posting the solution...

I think the solution is
$485.$
Am I correct?? ;)
Last edited by Labib on Thu Jan 06, 2011 11:53 pm, edited 1 time in total.
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Re: Find the last zeros of $2011!$

Unread post by Moon » Thu Jan 06, 2011 10:53 pm

Full solution please :)
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Avik Roy
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Re: Find the last zeros of $2011!$

Unread post by Avik Roy » Thu Jan 06, 2011 10:58 pm

I think we should be in hard line regarding 'not posting full solution'...this syndrome doesnt seem to cure...
Moon, u should post a "Almost Zero Tolerance" announcement regarding this
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor
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Labib
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Re: Find the last zeros of $2011!$

Unread post by Labib » Thu Jan 06, 2011 11:48 pm

Avik da, actually I post full solutions. But Moon vai said that it was for the juniors so I left it for them. anyway, here's the full solution::
$ \left \lfloor \frac {2011}{5}\right \rfloor + \left \lfloor \frac {2011}{25} \right \rfloor + \left \lfloor \frac {2011}{625} \right \rfloor =402+80+3=485 $

It's the power of $5$ in $2011!$.

$2$'s power is more than $5$'s for sure. so the number of $0$ is $485$..

(oops!!) :D
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