Find a positive integer $x$, when divided by 2 the remainder is 1, when divided by 5 the remainder is 2, when divided by 7 the remainder is 3, when divided by 9 the remainder is 4.

Ask for the original number. (Trans: Find a positive integer $x$, when divided by 2 the remainder is 1, when divided by 5 the remainder is 2, when divided by 7 the remainder is 3, when divided by 9 the remainder is 4.)
Solve According to the problem, we have
$$\begin{array}{lll}
x \equiv 1 & (\bmod 2), & x \equiv 2 \quad(\bmod 5) \\
x \equiv 3 & (\bmod 7), & x \equiv 4 \quad(\bmod 9)
\end{array}$$

In Theorem 1, take $m_{1}=2, m_{2}=5, m_{3}=7, m_{4}=9, b_{1}=1, b_{2}$ ? $=2, b_{3}=3, b_{4}=4$, at this point we have
$$\begin{array}{l}
M=2 \times 5 \times 7 \times 9=630 \\
M_{1}=\frac{630}{2}=315 \\
M_{2}=\frac{630}{5}=126 \\
M_{3}=\frac{630}{7}=90 \\
M_{4}=\frac{630}{9}=70
\end{array}$$

Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\operatorname{mod} 2)$, then we have $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 315 M_{1}^{\prime} \equiv M_{1}^{\prime}(\bmod 2)$, so we get $M_{1}^{\prime}=1$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 5)$, then we have $1 \equiv M_{2}^{\prime} M_{2} \equiv$ $126 M_{2}^{\prime} \equiv M_{2}^{\prime}(\bmod 5)$, so we get $M_{2}^{\prime}=1$. Let $M_{3}^{\prime}$ be a positive integer that satisfies $M_{3}^{\prime} M_{3} \equiv 1(\bmod 7)$, then we have $1 \equiv M_{3}^{\prime} M_{3} \equiv 90 M_{3}^{\prime} \equiv$ $6 M_{3}^{\prime}(\bmod 7)$, so we get $M_{3}^{\prime}=6$. Let $M_{4}^{\prime}$ be a positive integer that satisfies $M_{4}^{\prime} M_{4} \equiv 1(\bmod 9)$, then we have $1 \equiv M_{4}^{\prime} M_{4} \equiv 70 M_{4}^{\prime} \equiv 7 M_{4}^{\prime}$ (mod 9), so we get $M_{4}^{\prime}=4$. Therefore, by (46) we get
$$\begin{aligned}
x & \equiv 315+2 \times 126+3 \times 6 \times 90+4 \times 4 \times 70 \\
& \equiv 157(\bmod 630)
\end{aligned}$$

Thus, we have $\square$
$$x=157+630 k, \quad k=0,1,2, \cdots$$