For example, 41987 can be written as a three-digit number $x y z$ in base $b$. If $x+y+z=1+9+8+7$, try to determine all possible values of $x$, $y$, $z$, and $b$.
$(1987$, Canadian Mathematics Competition)
Analysis: It is easy to know that $x b^{2}+y b+z=1987, x+y+z=25$, thus,
$$
x\left(b^{2}-1\right)+y(b-1)=1962 \text {, }
$$

which means
$$
\begin{array}{l}
(b-1)[(b+1) x+y]=1962 \\
=2 \times 3^{2} \times 109 .
\end{array}
$$

From $b>10$ we know $b-1>9$;
From $1962 \geqslant b^{2}-1$ we know $b \leqslant \sqrt{1563}<45$.
Therefore, $3<b-1<44$.
Since $1962=2 \times 3^{2} \times 109$ has 12 positive divisors, which are
$$
1,2,3,6,9,18,109,218,327,654,981 \text {, }
$$
1962.
Thus, $b-1=18, b=19$.
From $1987=5 \times 19^{2}+9 \times 19+11$ we know $x=5, y=9, z=11$.