$1 \cdot 95$ Let $m$ and $n$ be known positive integers, $m$ has $d$ digits when represented in decimal, where $d \leqslant n$. Find the sum of all digits in the decimal representation of $\left(10^{n}-1\right) m$.
[Solution] Let $f(x)$ denote the sum of the digits of the positive integer $x$, and let $m=\overline{a_{1} a_{2} \cdots a_{d} 00 \cdots 0}, a_{d} \neq 0$.

Write $\left(10^{n}-1\right) m=m \cdot 10^{n}-m$ as a subtraction in vertical form, and note that $d \leqslant n$, we have $a_{1} a_{2} \cdots a_{d} 00 \cdots 000 \cdots 000 \cdots 0$

Therefore,
$$\begin{aligned}
f\left(\left(10^{n}-1\right) m\right) & =f\left(\overline{a_{1} a_{2} \cdots a_{d}} \cdot 10^{n}-\overline{a_{1} a_{2} \cdots a_{d}}\right) \\
& =f(m)-1+(9 n+1)-f(m) \\
& =9 n
\end{aligned}$$

Thus, the sum of the digits sought is $9 n$.