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There are floor(n/5) + floor(n/25) + floor(n/125)+... zeroes at the end of n! Since 1/5 + 1/25 + 1/125 + ... = 1/4, we would expect (4n!) to be a very good lower bound for our answer. Also, the number of zeroes in successive values of n! only increase when n is a multiple of five. Using those two facts, we can create a starting value that requires very few checks even when n is very large.
Manasa and Factorials
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There are floor(n/5) + floor(n/25) + floor(n/125)+... zeroes at the end of n! Since 1/5 + 1/25 + 1/125 + ... = 1/4, we would expect (4n!) to be a very good lower bound for our answer. Also, the number of zeroes in successive values of n! only increase when n is a multiple of five. Using those two facts, we can create a starting value that requires very few checks even when n is very large.
your code is so clean and the explanation is so clear. you re so intelligent
Beautiful code!
I also want to add this link for those interested in the idea behind it.
https://en.wikipedia.org/wiki/Trailing_zero