Example 4 Find the greatest common divisor of 6731 and 2809
Solve:
\begin{aligned}
6731 & =2809 \times 2+1113 \\
2809 & =1113 \times 2+583 \\
1113 & =583 \times 1+530 \\
583 & =530+53 \\
530 & =53 \times 10+0
\end{aligned}

So $(6731,2809)=53$. For convenience in writing, the series of calculations above can be abbreviated as follows:
2 \begin{tabular}{|r|r|r}
6731 & 2809 & 2 \\
5618 & 2226 \\
\hline 1113 & 583 & 1 \\
583 & 530 \\
\hline 530 & 53 & \\
530 & $(0)$ & \\
\hline 0 & 53 &.
\end{tabular}

Let $n \geqslant 3$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be positive integers. Here we introduce a method to find the greatest common divisor of these $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$: We first find the greatest common divisor of $a_{1}$ and $a_{2}$, and if the greatest common divisor of $a_{1}$ and $a_{2}$ is $b_{1}$. Then we find the greatest common divisor of $b_{1}$ and $a_{3}$, and if the greatest common divisor of $b_{1}$ and $a_{3}$ is $b_{2}$, then $b_{2}$ is the greatest common divisor of $a_{1}, a_{2}$, and $a_{3}$. When $n \geqslant 4$, we find the greatest common divisor of $b_{2}$ and $a_{4}$, and if the greatest common divisor of $b_{2}$ and $a_{4}$ is $b_{3}$, then $b_{3}$ is the greatest common divisor of $a_{1}, a_{2}, a_{3}$, and $a_{4}$. When $n \geqslant 5$, we find the greatest common divisor of $b_{3}$ and $a_{5}$, $\cdots$.