Example 13: When divided by 11, the remainder is 3; when divided by 12, the remainder is 2; when divided by 13, the remainder is 1. What is the original number? (Translation: Find a positive integer $x$ such that when divided by 11, the remainder is 3; when divided by 12, the remainder is 2; when divided by 13, the remainder is 1, )
According to the problem, we have
$$x \equiv 3(\bmod 11), \quad x \equiv 2(\bmod 12), \quad x \equiv 1(\bmod 13)$$

In Theorem 1, let $m_{1}=11, m_{2}=12, m_{3}=13, b_{1}=3, \dot{b}_{2}=2$, $b_{3}=1$, then we have
$$\begin{array}{l}
M=11 \times 12 \times 13=1716 \\
M_{1}=\frac{1716}{11}=156 \\
M_{2}=\frac{1716}{12}=143 \\
M_{3}=\frac{1716}{13}=132
\end{array}$$

Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\bmod 11)$, then $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 156 M_{1}^{\prime} \equiv 2 M_{1}^{\prime}(\bmod 11)$, hence $M_{1}^{\prime}=6$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 12)$, then $1 \equiv M_{2}^{\prime} M_{2} \equiv 143$ $M_{2}^{\prime} \equiv 11 M_{2}^{\prime}(\bmod 12)$, so we get $M_{2}^{\prime}=11$. Let $M_{3}^{\prime}$ be a positive integer that satisfies $M_{3}^{\prime} M_{3} \equiv 1(\bmod 13)$, then $1 \equiv M_{3}^{\prime} M_{3} \equiv 132 M_{3}^{\prime}$ $\equiv 2 M_{3}^{\prime}(\bmod 13)$, so we get $M_{3}^{\prime}=7$. Therefore, by equation (46) we get
$$\begin{aligned}
x & \equiv 3 \times 6 \times 156+2 \times 11 \times 143+7 \times 132 \\
\equiv & 14(\bmod 1716), \text { i.e., } \\
& x=14+1716 k, \quad k=0,1,2, \cdots .
\end{aligned}$$