Example 3. Suppose 1987 can be represented 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 values of $x$, $y$, $z$, and $b$. (Canadian 87 Competition Question)
Given the problem, we have
$$
\left\{\begin{array}{ll}
x b^{2}+y b+z=1987(x \geqslant 1), & (1) \\
x+y+z=25 & (2)
\end{array}\right.
$$
Subtracting (2) from (1) gives
$$
(b-1)[(b+1) x+y]=1962 \text {, }
$$

Thus, $(b-1) 1962=2 \times 9 \times 10$.
$$
\text { Also, } b^{3}>1987, b^{2}<1987 \text {, }
$$

Therefore, $12<b<45$. It follows that,
$$
b-1=2 \times 9=18 \text {, i.e., } b=19 \text {. }
$$

From the calculation
$$
19 \left\lvert\, \frac{1987}{19 \left\lvert\, \frac{104 \cdots 11}{5 \cdots 9}\right.}\right.
$$

we get $1987=5 \times 19^{2}+9 \times 19+11$.
Thus, $x=5, y=9, z=11, b=19$.
This problem is solved in base $b=19$, so $1 \leqslant x$, $y, z \leqslant 19$. Therefore, $z=11$ is completely consistent with the problem. The trick in this problem is that, after finding $b=19$, the positive integer solution to $19^{2} x+19 y+z=1987$ is directly obtained using the calculation, which is a unique method.