2018 0's

The number of trailing zeros of $N!$ is given by $z(N) = \displaystyle \sum_{k=1}^\infty \left \lfloor \frac N{5^k} \right \rfloor$ . Therefore, $\displaystyle \sum_{\color{#3D99F6}k=1}^\infty \left \lfloor \frac {5n}{5^k} \right \rfloor = \sum_{\color{#D61F06}k=0}^\infty \left \lfloor \frac n{5^k} \right \rfloor = 2018$ . We can estimate the value of $n$ by $\displaystyle \sum_{k=0}^\infty \frac n{5^k} = 2018$ $\implies \dfrac n{1-\frac 15} = 2018$ $\implies \dfrac 54 n = 2018$ $\implies n \approx 1614.4$ $\implies n \ge 1615$ . Let $n=1615+a$ . Then we have:

$\begin{aligned} \sum_{k=0}^\infty \left \lfloor \frac n{5^k} \right \rfloor & = 2018 \\ n + \left \lfloor \frac n5 \right \rfloor + \left \lfloor \frac n{25} \right \rfloor + \left \lfloor \frac n{125} \right \rfloor + \left \lfloor \frac n{625} \right \rfloor & = 2018 \\ 1615 + a + \left \lfloor \frac {1615 + a}5 \right \rfloor + \left \lfloor \frac {1615 + a}{25} \right \rfloor + \left \lfloor \frac {1615 + a}{125} \right \rfloor + \left \lfloor \frac {1615 + a}{625} \right \rfloor & = 2018 \\ 1615 + a + 323 + \left \lfloor \frac a5 \right \rfloor + 64 + \left \lfloor \frac {15+a}{25} \right \rfloor + 12 + \left \lfloor \frac {115 + a}{125} \right \rfloor + 2 + \left \lfloor \frac {365 + a}{625} \right \rfloor & = 2018 \\ 2016 + a + \left \lfloor \frac a5 \right \rfloor + \left \lfloor \frac {15+a}{25} \right \rfloor + \left \lfloor \frac {115 + a}{125} \right \rfloor + \left \lfloor \frac {365 + a}{625} \right \rfloor & = 2018 \\ \implies a & = 2 \end{aligned}$

Therefore, $n = 1615 + 2 = \boxed{1617}$ .

Suppose that $(5n)! = k!$ where $k$ is any natural number. We are given that $k!$ has $2018 = e$ zeros. We can make notice that $k > 4e \implies k > 4\cdot 2018 \implies k > 8072$ Since $k$ is the multiple of $5$ so $8075 \leq k \leq 8085$ or $\small 8075\leq k \leq 1617\cdot 5\implies k! \leq (5\cdot 1617)! = (5n)!$ . Therefore the required value of $\boxed{n=1617}$ .

No. of zeroes at the end of $5n! =$ numner of 5's and its multiples = $\lfloor \frac{5n}{5} \rfloor + \lfloor \frac{5n}{5^2} \rfloor + \cdots + \lfloor \frac{5n}{5^k} \rfloor = 5n \times \sum^1_k \frac{1}{5^k} =2018$

Set $k = \inf$ and slove to find the value of $n$ .