9. Try to solve:
(i) $\left\{\begin{array}{l}x \equiv 2 \quad(\bmod 7), \\ x \equiv 5 \quad(\bmod 9), \\ x \equiv 11 \quad(\bmod 15) .\end{array}\right.$
(ii) There is a number whose total is unknown; when reduced by multiples of five, there is no remainder; when reduced by multiples of seven hundred and fifteen, there is a remainder of ten; when reduced by multiples of two hundred and forty-seven, there is a remainder of one hundred and forty; when reduced by multiples of three hundred and ninety-one, there is a remainder of two hundred and forty-five; when reduced by multiples of one hundred and eighty-seven, there is a remainder of one hundred and nine. What is the total number? (Huang Zongxian: "General Solution of the Method of Seeking One," answer: 10020.)
9.
(i) Solution: Since $(7,9)=1,(7,15)=1,(9,15)=3,11-5=6$, and $3 \mid(11-5)$, by problem 8(i, ii), we know the system of congruences has a solution and is equivalent to
$$\left\{\begin{array}{l}
x \equiv 2 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9) \\
x \equiv 11 \equiv 1 \quad(\bmod 5)
\end{array}\right.$$

Solving the above using the Chinese Remainder Theorem:
$$\begin{array}{c}
b_{1}=2, \quad b_{2}=5, \quad b_{3}=1 \\
m_{1}=7, \quad m_{2}=9, \quad m_{3}=5 \\
m=m_{1} \cdot m_{2} \cdot m_{3}=7 \times 9 \times 5=315
\end{array}$$
$$M_{1}=\frac{315}{7}=45, \quad M_{2}=\frac{315}{9}=35, \quad M_{3}=\frac{315}{5}=63$$

From $45 M_{1}^{\prime} \equiv 1(\bmod 7)$, we get $M_{1}^{\prime}=-2$.
From $35 M_{2}^{\prime} \equiv 1(\bmod 9)$, we get $M_{2}^{\prime}=-1$.
From $63 M_{3}^{\prime} \equiv 1(\bmod 5)$, we get $M_{3}^{\prime}=2$.
Thus, the solution is
$$\begin{array}{c}
x \equiv 2 \times 45 \times(-2)+5 \times 35 \times(-1)+1 \times 63 \\
\times 2 \equiv-229 \equiv 86(\bmod 315) .
\end{array}$$
(ii) Solution: Let the total number be $x$. According to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10 \quad(\bmod 715) \\
x \equiv 140 \quad(\bmod 247) \\
x \equiv 245 \quad(\bmod 391) \\
x \equiv 109 \quad(\bmod 187)
\end{array}\right.$$

First, check if there is a solution. Since $715=5 \times 11 \times 13, 247=13 \times 19, 391=17 \times 23, 187=11 \times 17$, we have $(5,715)=5$, and $5 \mid(10-0)$; $(715,247)=13$, and $13 \mid(140-10)$; $(715,187)=11$, and $11 \mid(109-10)$; $(391,187)=17$, and $17 \mid(245-109)$; thus, by problem 8(i), the system of congruences has a solution. By problem 8(ii), appropriately choose the moduli so that the original system of congruences is equivalent to
$$\left\{\begin{array}{l}
x \equiv 0 \quad(\bmod 5) \\
x \equiv 10(\bmod 11 \times 13) \\
x \equiv 140 \equiv 7(\bmod 19) \\
x \equiv 245 \equiv 15(\bmod 23) \\
x \equiv 109 \equiv 7(\bmod 17)
\end{array}\right.$$

The moduli in the above system of congruences are pairwise coprime, so we can use the Chinese Remainder Theorem to solve it.
From $\square$
$$\begin{array}{c}
1062347 M_{1}^{\prime} \equiv 2 M_{1}^{\prime} \equiv 1 \quad(\bmod 5), \text { we get } M_{1}^{\prime}=3 . \\
37145 M_{2}^{\prime} \equiv 108 M_{2}^{\prime} \equiv 1 \quad(\bmod 11 \times 13) \\
\text { and } 11 \times 13 \equiv 143
\end{array}$$

Since

we have
$$\begin{array}{l}
143=108+35, \quad 108=3 \times 35+3, \\
35=11 \times 3+2, \quad 3=2+1,
\end{array}$$
$$\begin{aligned}
1 & =3-2=3-(35-11 \times 3)=12 \times 3-35 \\
& =12 \times(108-3 \times 35)-35 \\
& =12 \times 108-37 \times 35 \\
& =12 \times 108-37 \times(143-108) \\
& =49 \times 108-37 \times 143 .
\end{aligned}$$

Thus, $\square$
$108 \times 49 \equiv 1(\bmod 143)$, we get $M_{2}^{\prime}=49$.
From $279565 M_{3}^{\prime} \equiv 18 M_{3}^{\prime} \equiv 1(\bmod 19)$, we get $M_{3}^{\prime}=-1$.
From $230945 M_{4}^{\prime} \equiv 2 M_{4}^{\prime} \equiv 1(\bmod 23)$, we get $M_{4}^{\prime}=-11$.
From $312455 M_{5}^{\prime} \equiv 12 M_{5}^{\prime} \equiv 1(\bmod 17)$, we get $M_{5}^{\prime}=-7$.
Finally, the solution is
$$\begin{aligned}
x \equiv 10 & \times 37145 \times 49+7 \times 279565 \times(-1) \\
& +15 \times 230945 \times(-11)+7 \times 312455 \\
& \times(-7) \equiv 18201050-1956955-38105925 \\
& -15310295 \equiv-37172125 \\
\equiv & 10020(\bmod 5311735) .
\end{aligned}$$

Answer: The smallest total number is 10020.
$$\begin{array}{l}
b_{1}=0, \quad b_{2}=10, \quad b_{3}=7, \quad b_{4}=15, \quad b_{5}=7, \\
m_{1}=5, \quad m_{2}=11 \times 13, \quad m_{3}=19, \quad m_{4}=23, \\
m_{5}=17, \quad m=5 \times 11 \times 13 \times 19 \times 23 \times 17 \\
=5311735, \\
M_{1}=\frac{5311735}{5}=1062347, \\
M_{2}=\frac{5311735}{11 \times 13}=37145, \\
M_{3}=\frac{5311735}{19}=279565, \\
M_{4}=\frac{5311735}{23}=230945, \\
M_{5}=\frac{5311735}{17}=312455.
\end{array}$$