A natural number n such that n ! ends in exactly 1000 zeros is

Option $\left(1\right)$

$n=4010$

$$\begin{gathered} \left[\frac{4010}{5}\right]+\left[\frac{4010}{25}\right]+\left[\frac{4010}{125}\right]+\left[\frac{4010}{625}\right]+\left[\frac{4010}{3125}\right] \\ =802+160+32+6+1 \\ =1001>1000 \end{gathered}$$

Option $\left(2\right)$

$$\begin{gathered} \left[\frac{4000}{5}\right]+\left[\frac{4000}{25}\right]+\left[\frac{4000}{125}\right]+\left[\frac{4000}{625}\right]+\left[\frac{4000}{3125}\right] \\ =800+160+32+6+1 \\ =999<1000 \end{gathered}$$

Option $\left(3\right)$

$$\begin{gathered} \left[\frac{4009}{5}\right]+\left[\frac{4009}{25}\right]+\left[\frac{4009}{125}\right]+\left[\frac{4009}{625}\right]+\left[\frac{4009}{3125}\right] \\ =801+160+32+6+1 \\ =1000 \end{gathered}$$

Option $\left(4\right)$

$$\begin{gathered} \left[\frac{4004}{5}\right]+\left[\frac{4004}{25}\right]+\left[\frac{4004}{125}\right]+\left[\frac{4004}{625}\right]+\left[\frac{4004}{3125}\right] \\ =800+160+32+6+1 \\ =999<1000 \end{gathered}$$