1. How many zeros does the number 1986! end with?
Solution. Let $1986!=2^{m} \cdot 5^{n} \cdot k$, where $k$ is not divisible by either 2 or 5. Since the number 2 appears more frequently than 5 among the numbers from 1 to 1986, it follows that $m>n$. Since $10=2 \cdot 5$, it follows that 1986 ends with $n$ zeros. Among the numbers $1,2,3, \ldots, 1986$, there are $\left[\frac{1986}{5}\right]=397$ numbers that end in $5,\left[\frac{1986}{25}\right]=79$ numbers divisible by $5^{2},\left[\frac{1986}{125}\right]=15$ numbers divisible by $5^{3},\left[\frac{1986}{625}\right]=3$ numbers divisible by $5^{4}$, and there are no numbers divisible by $5^{5}$. Therefore, the number of zeros in 1986! is

$$
n=397+79+15+3=494
$$