6. Solve the following problems: (Yang Hui: "Xu Gu Zhai Qi Suan Fa" (1275))
(i) When divided by 7, the remainder is 1; when divided by 8, the remainder is 2; when divided by 9, the remainder is 4. What is the original number?
(ii) 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 5. What is the original number?
(iii) When divided by 11, the remainder is 3; when divided by 72, the remainder is 2; when divided by 13, the remainder is 1. What is the original number?
6.
(i) Solution: Let the number be $x$, then according to the problem, we have

Here
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By
$$\left.\begin{array}{c}
\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 7), \\
x \equiv 2 \quad(\bmod 8), \\
x \equiv 4 \quad(\bmod 9) .
\end{array}\right. \\
b_{1}=1, \quad b_{2}=2, \quad b_{3}=4, \\
m_{1}=7, \quad m_{2}=8, \quad m_{3}=9, \\
m=7 \times 8 \times 9=504,
\end{array}\right\}$$
Therefore,
$$\begin{array}{c}
x \equiv 1 \times 72 \times 4+2 \times 63 \times(-1)+4 \times 56 \\
\times(-4) \equiv-734 \equiv 274(\bmod 504)
\end{array}$$
(ii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{l}
x \equiv 1 \quad(\bmod 2) \\
x \equiv 2 \quad(\bmod 5) \\
x \equiv 3 \quad(\bmod 7) \\
x \equiv 5 \quad(\bmod 9)
\end{array}\right.$$

Here
$$\begin{array}{c}
b_{1}=1, \quad b_{2}=2, \quad b_{3}=3, \quad b_{4}=5 \\
m_{1}=2, \quad m_{2}=5, \quad m_{3}=7, \quad m_{4}=9 \\
m=m_{1} \cdot m_{2} \cdot m_{3} \cdot m_{4}=2 \times 5 \times 7 \times 9=630 \\
M_{1}=\frac{630}{2}=315, \quad M_{2}=\frac{630}{5}=126 \\
M_{3}=\frac{630}{7}=90, \quad M_{4}=\frac{630}{9}=70
\end{array}$$
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By
By
By
Therefore,
$$\begin{array}{l} 
315 M_{1}^{\prime} \equiv 1 \quad(\bmod 2), \text { get } M_{1}^{\prime}=1 \\
126 M_{2}^{\prime} \equiv 1 \quad(\bmod 5), \text { get } M_{2}^{\prime}=1 \\
90 M_{3}^{\prime} \equiv 1 \quad(\bmod 7), \\
70 M_{4}^{\prime} \equiv 1 \quad(\bmod 9), M_{3}^{\prime}=-1 \\
x=315+2 \times 126+3 \times 90 \times(-1)+5 \\
\times 70 \times 4 \equiv 1697 \equiv 437(\bmod 630)
\end{array}$$
(iii) Solution: Let the number be $x$, according to the problem, we have
$$\left\{\begin{array}{ll}
x \equiv 3 & (\bmod 11) \\
x \equiv 2 & (\bmod 72) \\
x \equiv 1 & (\bmod 13)
\end{array}\right.$$

Here
$$\begin{array}{c}
b_{1}=3, \quad b_{2}=2, \quad b_{3}=1, \\
m_{1}=11, \quad m_{2}=72, \quad m_{3}=13, \\
m=m_{1} \cdot m_{2} \cdot m_{3}=11 \times 72 \times 13=10296, \\
M_{1}=\frac{10296}{11}=936, \quad M_{2}=\frac{10296}{72}=143, \\
M_{3}=\frac{10296}{13}=792 . \\
936 M_{1}^{\prime} \equiv 1 \quad(\bmod 11), \text { get } M_{1}^{\prime}=1 . \\
143 M_{2}^{\prime} \equiv 1 \quad(\bmod 72), \text { get } M_{2}^{\prime}=-1 . \\
792 M_{3}^{\prime} \equiv 1 \quad(\bmod 13), \text { get } M_{3}^{\prime}=-1 .
\end{array}$$

Therefore,
$$\begin{array}{l} 
x \equiv 3 \times 936+2 \times 143 \times(-1)+1 \times 792 \\
\times(-1) \equiv 1730(\bmod 10296)
\end{array}$$