jmc

To find the number of terminal zeros, we must find the number of products $2\times 5$ in $236!$. Since there are more factors of 2 than factors of 5, we can get our answer by finding the largest power of 5 that divides $236!$. Every multiple of 5 less than 236 gives a factor of 5, each multiple of 25 gives an additional factor of 5, and each multiple of 125 gives a third factor of 5. Therefore, the number of factors of 5 in $236!$ is $\lfloor \frac{236}{5}\rfloor +\lfloor \frac{236}{25}\rfloor +\lfloor \frac{236}{125}\rfloor =47+9+1=57$. The highest power of 5 that divides $236!$ is $5^{57}$ so $236!$ ends in $\boxed{57}$ zeros.