4- 251 Suppose 1987 can be written as a three-digit number $x y z$ in base $b$ and $x+y+z=1+9+8+7$, try to determine all $x, y, z$ and $b$.


［Solution］ From the given, we have
$$
x b^{2}+y b+z=1987(x \geqslant 1)
$$

and $b^{3}>1987, b^{2}<1987$. Therefore,
$$
12<b<45 \text {. }
$$

Also, from the given, we know $x+y+z=25$.
Subtracting (2) from (1) gives $(b-1)(b x+x+y)=1962$.

Thus, $b-1$ can divide 1962. Considering $1962=2 \times 9 \times 109$,
and $12<b<45$,
we find $b=19$.
Therefore, $1987=5 \times 19^{2}+9 \times 19+11$,
which means $x=5, y=9, z=11, b=19$.