Example 11 Find the integer solution of $1296 x \equiv 1125(\bmod 1935)$.
Since $(1296,1935)=9,9 \mid 1125$, from the original equation we get
$$144 x \equiv 125(\bmod 215)$$

Since $(144,215)=1$, by Lemma 12 we know that $144 x \equiv 125(\bmod 215)$ has integer solutions. Since $215=144+71,144=71 \times 2+2$, and $71=2 \times 35+1$, we get
$$\begin{aligned}
1= & 71-2 \times 35=71-(144-71 \times 2) \times 35 \\
= & -144 \times 35+71 \times 71=-144 \times 35+71 \\
& \times(215-144)=-106 \times 144+71 \times 215
\end{aligned}$$

So we have
$$-106 \times 144 \equiv 1(\bmod 215)$$

By equation (39) and Lemma 5, we have
$$-106 \times 144 x \equiv-106 \times 125(\bmod 215)$$

Since $-106 \times 125=80-215 \times 62$, we get
$$-125 \times 106 \equiv 80(\bmod 215)$$

From equations (40) to (42), we get $x \equiv 80(\bmod 215)$. Since $1935=215 \times 9$ and $80+215=295,80+2 \times 215=510,80+3 \times 215=725,80+4 \times 215=940,80+5 \times 215=1155,80+6 \times 215=1370,80+7 \times 215=1585,80+8 \times 215=1800$, the 9 different solutions to $1296 x \equiv 1125(\bmod 1935)$ are
$$\begin{aligned}
x \equiv & 80,295,510,725,940,1155,1370,1585 \\
& 1800(\bmod 1935)
\end{aligned}$$