Example 4 The sequence of positive integers $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+b_{n}(n \geqslant 1) \text {, }
$$

where $b_{n}$ is obtained by reversing the digits of $a_{n}$ (the leading digit of $b_{n}$ can be zero), for example, when $a_{1}=170$, we have $a_{2}=241, a_{3}=383, a_{4}=766$, etc. Question: Can $a_{7}$ be a prime number ${ }^{[2]}$
(55th Czech and Slovak Mathematical Olympiad)
【Analysis】First, we conduct a simple mathematical experiment, taking $a_{1}=1$, then
$$
\begin{array}{l}
a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, \\
a_{6}=77, a_{7}=847=7 \times 11^{2} .
\end{array}
$$

Conjecture: $a_{7}$ is not a prime number and contains a certain special prime factor.

The prime number 3, which is unrelated to the reversal of digits, is obviously not that special prime factor. It is easy to know that the prime number 7 has a very small relationship with the reversal of digits, while the prime number 11 has a relatively larger relationship with the reversal of digits.
It is easy to prove: if
$$
11\left|a_{\mathrm{n}} \Rightarrow 11\right| b_{\mathrm{n}} \Rightarrow 11 \mid a_{\mathrm{n}+1} \text {. }
$$

On the other hand, if $a_{n}$ has an even number of digits, i.e.,
$$
a_{n}=\overline{x_{2 m-1} x_{2 m-2} \cdots x_{1} x_{0}} \text {, }
$$

then
$$
\begin{aligned}
a_{n} \equiv & x_{0}-x_{1}+x_{2}-x_{3}+\cdots+x_{2 m-2}-x_{2 m-1}(\bmod 11), \\
b_{n} \equiv & x_{2 m-1}-x_{2 m-2}+x_{2 m-3}-x_{2 m-4}+\cdots+ \\
& x_{1}-x_{0}(\bmod 11) .
\end{aligned}
$$

Thus, $a_{n+1}=a_{n}+b_{n} \equiv 0(\bmod 11)$.
If $a_{1}$ has an even number of digits, then $11 \mid a_{2}$, hence $11 \mid a_{7}$.
If $a_{1}$ has an odd number of digits, we will next prove that at least one of $a_{2}, a_{3}$, $\cdots, a_{6}$ has an even number of digits, hence $11 \mid a_{7}$. Otherwise, assume that $a_{2}, a_{3}, \cdots, a_{6}$ all have an odd number of digits.
Let $a_{1}=\overline{x_{2 k} x_{2 k-1} \cdots x_{1} x_{0}}$. Then
$$
b_{1}=\overline{x_{0} x_{1} \cdots x_{2 k}} .
$$

According to the assumption, $a_{2}, a_{3}, \cdots, a_{6}$ must all be $2 k+1$ digits. Thus,
$$
\left\{\begin{array}{l}
a_{2} \text { has } 2 k+1 \text { digits } \Rightarrow x_{0}+x_{2 k}<10, \\
a_{2} \text { 's last digit }=x_{0}+x_{2 k}, \\
a_{2} \text { 's first digit } \\
=x_{0}+x_{2 k}+0 \text { or } x_{0}+x_{2 k}+1 .
\end{array}\right.
$$

Similarly, $\left\{\begin{array}{l}a_{3} \text { has } 2 k+1 \text { digits } \Rightarrow 2\left(x_{0}+x_{2 k}\right)<10, \\ a_{3} \text { 's last digit } \geqslant 2\left(x_{0}+x_{2 k}\right), \\ a_{3} \text { 's first digit } \geqslant 2\left(x_{0}+x_{2 k}\right) ;\end{array}\right.$ $\left\{\begin{array}{l}a_{4} \text { has } 2 k+1 \text { digits } \Rightarrow 4\left(x_{0}+x_{2 k}\right)<10, \\ a_{4} \text { 's last digit } \geqslant 4\left(x_{0}+x_{2 k}\right), \\ a_{4} \text { 's first digit } \geqslant 4\left(x_{0}+x_{2 k}\right) ;\end{array}\right.$ $\left\{\begin{array}{l}a_{5} \text { has } 2 k+1 \text { digits } \Rightarrow 8\left(x_{0}+x_{2 k}\right)<10, \\ a_{5} \text { 's last digit } \geqslant 8\left(x_{0}+x_{2 k}\right), \\ a_{5} \text { 's first digit } \geqslant 8\left(x_{0}+x_{2 k}\right) ;\end{array}\right.$
$a_{6}$ has $2 k+1$ digits $\Rightarrow 16\left(x_{0}+x_{2 k}\right)<10$, which is clearly impossible.

Therefore, the assumption is not valid, i.e., at least one of $a_{2}, a_{3}, \cdots, a_{6}$ has an even number of digits, hence $11 \mid a_{7}$.
If $a_{7}$ is a prime number, then $a_{7}=11$.
In this case, $a_{6}=10, a_{5}=5$, and it is impossible to determine $a_{4}, a_{3}, a_{2}, a_{1}$.
Therefore, $a_{7} \neq 11$, i.e., $a_{7}$ is not a prime number.