9. Use the Euclidean algorithm to find the greatest common divisor:
(i) 51425, 13310.
(ii) $353430, 530145, 165186$.
(iii) $81719, 52003, 33649, 30107$.
9
(i) Solution: Since
1 \begin{tabular}{r|r|r}
51425 & 13310 \\
39930 & 11495 \\
\hline 11495 & 1815 & 6 \\
10890 & 1815 \\
\hline 605 & 0
\end{tabular}

Therefore, $(51425,13310)=605$.
(ii) Solution: Since

We get
$$(353430,530145)=176715$$

Also,
$14 \left\lvert\,$\begin{tabular}{r|r|}
176715 & 165186 \\
165186 & 161406 \\
\hline 11529 & 3780 \\
11340 & 3780 \\
\hline 189 & 0
\end{tabular}\right.,

We get
$(176715,165186)=189$.
Therefore, $(353430,530145,165186)=189$.
(iii) Solution: Since
$1\left|\begin{array}{r|r|r}81719 & 52003 \\ 52003 & 29716 \\ \hline 29716 & 22287 \\ 22287 & 1 \\ \hline 7429 & 0\end{array}\right|$

We get $(81719,52003)=7429$
Since
\begin{tabular}{|c|c|c|}
\hline & 33649 & 7429 \\
\hline & 29716 & 3933 \\
\hline 1 & 3933 & 3496 \\
\hline & 3496 & 3496 \\
\hline 8 & 437 & 0 \\
\hline
\end{tabular}

We get
$(33649,7429)=437$.
Since
\begin{tabular}{|c|c|c|c|}
\hline & 30107 & 437 & 68 \\
\hline & 29716 & 391 & \\
\hline 1 & 391 & 46 & 8 \\
\hline & 368 & 46 & \\
\hline 2 & 23 & 0 & \\
\hline
\end{tabular}

We get
$(30107,437)=23$.
Therefore, $(81719,52003,33649,30107)=23$.