Question Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$
$(2007$, National High School Mathematics League Jiangxi Province Preliminary Contest)
Another solution: Considering that $a_{n}$ is not continuous in decimal, and the digits of $a_{n}$ are all no greater than 3 (i.e., $0, 1, 2, 3$), then the form of $a_{n}$ is a sequence of consecutive integers in quaternary, thus simplifying the problem.

Let the set $\left\{\left(a_{n}\right)_{4}\right\}=\{$ consecutive positive integers in quaternary $\}$ $\left((x)_{k}\right.$ denotes $x$ as a $k$-ary number, for example, $(11)_{4}=$ (5) 10 ). Therefore,
$$
\begin{array}{l}
\left(a_{n}\right)_{4}=\left(a_{n-1}\right)_{4}+(1)_{10} \\
=\left(a_{n-2}\right)_{4}+(2)_{10}=\cdots=(n)_{10} .
\end{array}
$$

When $n=2007$, we have
$$
\left(a_{2007}\right)_{4}=(2007)_{10} \text {. }
$$

Therefore, $a_{2007}=133113$.
Table 1 illustrates this.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $a_{1}$ & $a_{2}$ & $a_{3}$ & $c_{4}$ & $a_{5}$ & $a_{6}$ & $a_{7}$ & $a_{8}$ & $a_{9}$ & $a_{10}$ & $1 \ldots$ & $a_{2007}$ \\
\hline Subscript & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\cdots$ & 2007 \\
\hline Quaternary & 1 & 2 & 3 & 10 & 11 & 12 & 13 & 20 & 21 & $n$ & $\cdots$ & 133113 \\
\hline Value of $a_{n}$ & 1 & 2 & 3 & 10 & 11 & 12 & 13 & 20 & 21 & 22 & $\cdots$ & 133113 \\
\hline
\end{tabular}