There are 10 total numbers between 1 and 52 which include 5 as a factor (5 which also include 2 as a factor to make a factor of 10 and 5 more to be multiplied with a bunch of other 2-factor numbers) so intuitively I was thinking there should be 10 zeroes. The two I missed are additional factors of 5, one each in 25 and 50.
And from this (or by other arguably more elegant means) one can get the related cute theorem: the number of times p divides the binomial coefficient (n choose r) is the number of carries that occur when adding r to n-r in base p.
In particular, if n = p^k then unless r=0 or r=n there is always at least one carry because n is "longer" than r and n-r, so all the binomial coefficients (p^k choose r) are multiples of p apart from (p^k choose 0) and (p^k choose p^k).
(You can use this sort of idea to understand why, if you write out many many rows of Pascal's triangle mod 2, you get a sort of Sierpinski gasket thing.)
There is a 5^12 and more than that for the 2s. So yes 12 zeros it will be as God intended.
https://www.wolframalpha.com/input?i=factors+of+806581751709...