Problem 10

10. How many zeros are there at the end of the number 1000!=(1)(2)(3)...(999)(1000)? How many for 1001!, 1002!, 1003!, 1004!, and 1005!?

Proof: Let $n=2^a5^bm$ with $2,5~\not |~m$. Then $n$ has $\text{min}\{a,b\}$ zeros at the end, as $10^c~|~n⇔2^c,5^c~|~n⇔c≤\text{min}\{a,b\}$. By Legendre's theorem for the power of $p$ in $n!$ $$v_p(n!)=\sum_{k=1} \lfloor n/p^k \rfloor$$ we see we will have to pay attention only to the power of $5$ in computing the number of zeros at the end. Use Legendre's theorem for this application on $1000$ to obtain $249$ zeros, and the same goes for $1001!,...,1004!$ as they don't have any greater powers of $5$, and $1005=3·5·67$ contributes one extra zero to $1005!$ for a total of $250$.