Give the number of zeros at the end of 1000!. 

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Evidently, this number $N$ is $\min \left(v_{2}(1000!), v_{5}(1000!)\right)$. The well-known Legendre's formula states that for a prime $p$, and an integer $n$,

$$
v_{p}(n!)=\sum_{i=1}^{+\infty}\left\lfloor\frac{n}{p^{i}}\right\rfloor
$$

the sum being zero from a certain rank (for example, for $i \geq n$). Thus, $v_{p}(n!)$ decreases as $p$ increases, and $N=v_{5}(1000!)=\left\lfloor\frac{1000}{5}\right\rfloor+\left\lfloor\frac{1000}{25}\right\rfloor+\left\lfloor\frac{1000}{125}\right\rfloor+\left\lfloor\frac{1000}{625}\right\rfloor$, the subsequent terms being zero. Therefore, $N=200+40+8+1=249$.