Example 3 Find a complete residue system modulo 11, such that each number leaves remainders of $1, -1, 1, -1$ when divided by $2, 3, 5, 7$ respectively.
Solve: In Theorem 2, take $m_{1}=2, m_{2}=3, m_{3}=5, m_{4}=7, m_{5}=11$, and $x^{(1)}=1, x^{(2)}=-1, x^{(3)}=1, x^{(4)}=-1$. Thus, by Theorem 2, when $x^{(5)}$ runs through a complete residue system modulo 11,
$$x=M_{1} M_{1}^{-1}-M_{2} M_{2}^{-1}+M_{3} M_{3}^{-1}-M_{4} M_{4}^{-1}+M_{5} M_{5}^{-1} x^{(5)}$$

gives the required complete residue system. Now, we find $M_{j}^{-1}(1 \leqslant j \leqslant 5)$. By $M_{1} \equiv$ $1(\bmod 2)$ we have
$$1 \equiv M_{1} M_{1}^{-1} \equiv M_{1}^{-1}(\bmod 2)$$

so we can take $M_{1}^{-1}=1$. By $M_{2} \equiv-1(\bmod 3)$ we have
$$1 \equiv M_{2} M_{2}^{-1} \equiv(-1) \cdot M_{2}^{-1}(\bmod 3)$$

so we can take $M_{2}^{-1}=-1$. By $M_{3} \equiv 2(\bmod 5)$ we have
$$1 \equiv M_{3} M_{3}^{-1} \equiv 2 M_{3}^{-1}(\bmod 5)$$

so we can take $M_{3}^{-1}=-2$. By $M_{4} \equiv 1(\bmod 7)$ we have
$$1 \equiv M_{4} M_{4}^{-1} \equiv M_{4}^{-1}(\bmod 7)$$

so we can take $M_{4}^{-1}=1$. By $M_{5} \equiv 1(\bmod 11)$ we have
$$1 \equiv M_{5} M_{5}^{-1} \equiv M_{5}^{-1}(\bmod 11)$$

so we can take $M_{5}^{-1}=1{ }^{\oplus}$. Thus, we get
$$\begin{aligned}
x= & 3 \cdot 5 \cdot 7 \cdot 11+2 \cdot 5 \cdot 7 \cdot 11+2 \cdot 3 \cdot 7 \cdot 11 \cdot(-2) \\
& -2 \cdot 3 \cdot 5 \cdot 11+2 \cdot 3 \cdot 5 \cdot 7 x^{(5)} \\
= & 1155+770-924-330+210 x^{(5)} \\
= & 671+210 x^{(5)}=210\left(x^{(5)}+3\right)+41
\end{aligned}$$

The smallest positive complete residue system modulo 11 with this property is
$$\begin{array}{c}
41,210+41,210 \cdot 2+41 \\
210 \cdot 3+41,210 \cdot 4+41,210 \cdot 5+41 \\
210 \cdot 6+41,210 \cdot 7+41,210 \cdot 8+41 \\
210 \cdot 9+41,210 \cdot 10+41
\end{array}$$