Promotion 1 Select $n$ $(1 \leqslant n \leqslant 10)$ consecutive digits $a_{1}, a_{2}, \cdots, a_{n}$ from the ten digits $0,1, \cdots, 9$, to form an $n$-digit number $P$, where, without loss of generality, let
$$
a_{1}=a_{2}-1=a_{3}-2=\cdots=a_{n}-n+1 \text {. }
$$

Arrange all $P$ in ascending order, find what $P_{m}$ equals.
Solution: First, we make the following stipulations:
For $x, y, k \in \mathbf{N} (k$ being the base), $k \geqslant 2$, if $(y)_{k}=x$, then $y=\varphi_{k}(x)$.

For any $P_{i}$, let $Q_{i}=P_{i}-\underbrace{a_{1} a_{1} \cdots a_{1}}_{n \uparrow}$. Thus, $\left\{Q_{i}\right\}$ satisfies the requirement of being a sequence of non-negative integers in base $n$. Hence,
$$
\left(Q_{i}\right)_{n}=i-1 \Rightarrow Q_{i}=\varphi_{n}(i-1) .
$$
$$
\begin{array}{l}
\text { Then } P_{m}=Q_{m}+\underbrace{a_{1} a_{1} \cdots a_{1}}_{n \uparrow} \\
=\varphi_{n}(m-1)+\underbrace{a_{1} a_{1} \cdots a_{1}}_{n \uparrow} .
\end{array}
$$