In a string composed of decimal digits, if it contains an even number of digit 0, it is considered a valid codeword. For example, 001230407986 is a valid codeword, while 209807004508 is an invalid codeword. Find the number of valid codewords of length $n$.
Let $a_{n}$ denote the number of valid codewords of length $n$. Since in a digit string of length 1, only the digit string 0 is an invalid codeword, we have $a_{1}=9$. A valid codeword of length $n$ can be obtained by adding a digit to a digit string of length $n-1$ in two ways.

The first method is to add a non-zero digit to the end of a valid codeword of length $n-1$, so the number of valid codewords of length $n$ obtained by this method is $9 a_{n-1}$.

The second method is to add the digit 0 to the end of an invalid codeword of length $n-1$. Since the number of invalid codewords of length $n-1$ is $10^{n-1}-a_{n-1}$, the number of valid codewords of length $n$ obtained by this method is $10^{n-1}-a_{n-1}$.
Therefore, we have
$$
a_{n}=9 a_{n-1}+\left(10^{n-1}-a_{n-1}\right)
$$
$$
=8 a_{n-1}+10^{n-1} \text {. }
$$

Rewriting the above equation, we get
$$
a_{n}-\frac{10^{n}}{2}=8\left(a_{n-1}-\frac{10^{n-1}}{2}\right)
$$

The sequence $\left\{a_{n}-\frac{10^{n}}{2}\right\}$ is a geometric sequence with a common ratio of 8 and the first term $a_{1}-\frac{10}{2}=4$. It is not difficult to find that
$$
a_{n}-\frac{10^{n}}{2}=4 \times 8^{n-1}=\frac{8^{n}}{2}, a_{n}=\frac{8^{n}+10^{n}}{2} .
$$

Therefore, the number of valid codewords of length $n$ is $\frac{1}{2}\left(8^{n}+10^{n}\right)$.
(Tian Zhengping, Department of Mathematics, Hangzhou Normal University, 310036)