Let $x$ be the number of soldiers we are looking for. According to the problem,

In Theorem 1, take $m_{1}=5, m_{2}=6, m_{3}=7, m_{4}=11, b_{1}=1$, $b_{2}=5, b_{3}=4, b_{4}=10$. Then we have
$$\begin{array}{l} M=5 \times 6 \times 7 \times 11=2310 \\ M_{1}=\frac{2310}{5}=462 \\ M_{2}=\frac{2310}{6}=385 \\ M_{3}=\frac{2310}{7}=330 \\ M_{4}=\frac{2310}{11}=210 \end{array}$$

Let $M_{1}^{\prime}$ be a positive integer that satisfies $M_{1}^{\prime} M_{1} \equiv 1(\bmod 5)$, then $1 \equiv$ $M_{1}^{\prime} M_{1} \equiv 462 M_{1}^{\prime} \equiv 2 M_{1}^{\prime}(\bmod 5)$, so we get $M_{1}^{\prime}=3$. Let $M_{2}^{\prime}$ be a positive integer that satisfies $M_{2}^{\prime} M_{2} \equiv 1(\bmod 6)$, then $1 \equiv M_{2}^{\prime} M_{2} \equiv$ $385 M_{2}^{\prime} \equiv M_{2}^{\prime}(\bmod 6)$, 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 $1 \equiv M_{3}^{\prime} M_{3} \equiv 330 M_{3}^{\prime} \equiv$ $M_{3}^{\prime}(\bmod 7)$, so we get $M_{3}^{\prime}=1$. Let $M_{4}^{\prime}$ be a positive integer that satisfies $M_{4}^{\prime} M_{4} \equiv 1 \quad(\bmod 11)$, then $1 \equiv M_{4}^{\prime} M_{4} \equiv 210 M_{4}^{\prime} \equiv M_{4}^{\prime} (\bmod$ 11). Therefore, by (46) we get
$$\begin{aligned} x & \equiv 3 \times 462+5 \times 385+4 \times 330+10 \times 210 \\ & \equiv 6731 \equiv 2111(\bmod 2310) \end{aligned}$$

Thus, we have
$$x=2111+2310 k, \quad k=0,1,2, \cdots$$
$$\begin{array}{l} x \equiv 1(\bmod 5), x \equiv 5(\bmod 6), \\ x \equiv 4 \quad(\bmod 7), \quad x \equiv 10 \quad(\bmod 11). \end{array}$$