$11 \cdot 31$ Suppose 1987 can be written as a three-digit number $\overline{x y z}$ in base $b$, and $x+y+z=$ $1+9+8+7$, try to determine all possible $x, y, z$ and $b$.
(19th Canadian Mathematics Competition, 1987)
［Solution］ From the problem, in base ten we have

then
$$
x b^{2}+y b+z=1987, \quad(x \geqslant 1)
$$
$$
b^{3}>1987, \quad b^{2} \leqslant 1987 .
$$

From this, we get $12<b<45$.
Furthermore, from the problem,
$$
x+y+z=1+9+8+7=25,
$$
(1) - (2) gives
$$
(b-1)(b x+x+y)=1962 .
$$

Thus, $b-1 \mid 1962=2 \cdot 9 \cdot 109$.
Since $\quad 11<b-1<44$.
Therefore, it can only be
$$
b-1=18, \quad b=19 .
$$

So the base $b$ in the problem is base 19.
Convert the decimal number 1987 to its base 19 representation:

then
$$
1987=5 \cdot 19^{2}+9 \cdot 19+11 \text {. }
$$

Therefore,
$$
x=5, y=9, z=11, b=19 .
$$