24[\frac {n}{5}]+[\frac {n}{5^2}]+[\frac {n}{5^3}]+.....[\frac {n}{5^n}]
9 Answers
24
1yeah.
basically we can take this till ∞ only adding lots of zeros wont help. :P
21E_{2}(n!) = \left[\frac{n}{2} \right] + \left[\frac{n}{2^{2}} \right]....
E_{5}(n!) = \left[\frac{n}{5} \right] + \left[\frac{n}{5^{2}} \right]....
THE NO OF ZEROES AT THE END OF n! is min(E_{5}(n!),E_{2}(n!))
24@ deepak.. E2>E5 always..so minimum of them will be E5
so no. of zeros =E5
62Dont go by this formula unless you have understood why...
Sorry for presuming this... but do read these :
http://targetiit.com/iit-jee-forum/posts/help-1697.html
http://targetiit.com/iit-jee-forum/posts/9th-january-2009-1549.html
If you have any doubts, feel free to get clarified :)
