Problem 1433

5.
(i) Solution: By the Chinese Remainder Theorem,
Given
Given
$$\begin{array}{c} b_{1}=3, \quad b_{2}=5, \quad m_{1}=7, \quad m_{2}=11, \\ m=m_{1} \cdot m_{2}=7 \times 11=77, \\ M_{1}=\frac{77}{7}=11, \quad M_{2}=\frac{77}{11}=7 . \\ 11 M_{1}^{\prime}=1 \quad(\bmod 7), \text { so } M_{1}^{\prime}=2 . \\ 7 M_{2}^{\prime}=1 \quad(\bmod 11), \text { so } M_{2}^{\prime}=8 . \end{array}$$

Therefore, the solution is
$$\begin{aligned} x & \equiv 3 \times 11 \times 2+5 \times 7 \times 8 \\ & \equiv 346 \equiv 38(\bmod 77) \end{aligned}$$
(ii) Solution: The moduli are pairwise coprime, so the Chinese Remainder Theorem can be used.
$$\begin{array}{c} b_{1}=2, \quad b_{2}=5, \quad b_{3}=4 \\ m_{1}=11, \quad m_{2}=7, \quad m_{3}=5 \\ m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 7 \times 5=385 \\ M_{1}=\frac{385}{11}=35, \quad M_{2}=\frac{385}{7}=55 \\ M_{3}=\frac{385}{5}=77 \end{array}$$

By
That is
$$35 M_{1}^{\prime} \equiv 1 \quad(\bmod 11)$$

So
$$\begin{array}{c} (11 \times 3+2) M_{1}^{\prime} \equiv 1 \quad(\bmod 11) \\ 2 M_{1}^{\prime} \equiv 1(\bmod 11) \end{array}$$
We get
$$M_{1}^{\prime}=-5$$

Similarly, by
$$55 M_{2}^{\prime} \equiv 1(\bmod 7) ， \quad \text { i.e., } 6 M_{2}^{\prime} \equiv 1(\bmod 7) ，$$

We get
$$M_{2}^{\prime}=-1$$

By
$$77 M_{3}^{\prime} \equiv 1(\bmod 5) ， \text { i.e., } 2 M_{3}^{\prime} \equiv 1(\bmod 5) \text { ， }$$
We get
$$M_{3}^{\prime}=3$$

By the Chinese Remainder Theorem, the solution is
$$\begin{aligned} x \equiv & 2 \times 35 \times(-5)+5 \times 55 \times(-1) \\ & +4 \times 77 \times 3 \equiv 299 \quad(\bmod 385) \end{aligned}$$
(iii) Solution: From the second and third equations, we get
$$x \equiv 3(\bmod 5), \quad x \equiv 5(\bmod 9)$$

Combining the first equation with the above two equations, we can use the Chinese Remainder Theorem to solve.
$$\begin{array}{c} b_{1}=1, \quad b_{2}=3, \quad b_{3}=5 \\ m_{1}=7, \quad m_{2}=5, \quad m_{3}=9 \\ m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 5 \times 9=315 \end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{5}=63, \quad M_{3}=\frac{315}{9}=35$$
By
$45 M_{1}^{\prime} \equiv 1 \quad(\bmod 7)$ , we get $M_{1}^{\prime}=-2$ .
By $\quad 63 M_{2}^{\prime} \equiv 1(\bmod 5)$ , we get $M_{2}^{\prime}=2$ .
By $\quad 35 M_{3}^{\prime} \equiv 1(\bmod 9)$ , we get $M_{3}^{\prime}=-1$ .
Therefore,
$$\begin{aligned} x \equiv & 1 \times 45 \times(-2)+3 \times 63 \times 2 \\ & +5 \times 35 \times(-1) \equiv 113(\bmod 315) \end{aligned}$$