4.8 There is a four-digit number less than 2000, which has exactly 14 positive divisors (including 1 and itself), and one of its prime divisors ends with the digit 1. Find this four-digit number.
(Shanghai Junior High School Mathematics Competition, 1984)
For natural numbers, $i=1,2, \cdots, k$. The number of divisors of $N$ is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)=14=2 \cdot 7=1 \cdot 14 .
$$

Since $\alpha_{i} \geqslant 1$, then $\alpha_{i}+1 \geqslant 2$. Therefore, $k \leqslant 2$.
When $k=1$, we have $\quad p_{1}^{13}\alpha_{2}$, then $\alpha_{1}=6, \alpha_{2}=1$.
So
$$
N=p_{1}^{6} \cdot p_{2} \text {. }
$$

Because
$$
p_{1}^{6} \cdot p_{2} < 2000 \text {. }
$$
Therefore
$$
p_{1}=2 \text {. }
$$

Thus
$$
N=2^{6} \cdot p_{2} \text {. }
$$

If $p_{2}=11$, then $N=2^{6} \cdot 11=704$ is not a four-digit number.
If $p_{2}=31$, then $N=2^{6} \cdot 31=1984$.
If $p_{2} \geqslant 41$, then $N \geqslant 2^{6} \cdot 41 > 2000$.
Therefore, the only four-digit number is 1984.