A number theory problem by Nikhil Raj

Relevant wiki: Trailing Number of Zeros

The formula for finding the number of trailing zeros in $n!$ is:

$\begin{aligned} f(n) &= \sum_{i \ = \ 1}^k \left\lfloor{\frac{n}{5^i}}\right\rfloor = \left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{5^2}}\right\rfloor+\left\lfloor{\frac{n}{5^3}}\right\rfloor+\dots+\left\lfloor{\frac{n}{5^k}}\right\rfloor, \end{aligned}$

where $k = \left\lfloor \log_5 n \right\rfloor .$

For $\displaystyle n = 5000$ , $\displaystyle k = 5$ :

$\begin{aligned} f(5000) & = \sum_{i \ = \ 1}^5 \left\lfloor{\frac{5000}{5^n}}\right\rfloor \\ \\ & = \left\lfloor{\frac{5000}{5^1}}\right\rfloor+\left\lfloor{\frac{5000}{5^2}}\right\rfloor+\left\lfloor{\frac{5000}{5^3}}\right\rfloor+\left\lfloor{\frac{5000}{5^4}}\right\rfloor+\left\lfloor{\frac{5000}{5^5}}\right\rfloor \\ \\ & = 1000 + 200 + 40 + 8 + 1 \\ & = 1249 \end{aligned}$