Example 10 Find the integer solution of $1215 x \equiv 560(\bmod 2755)$.
Solving: Since $(1215,2755)=5,5 \mid 560$, from the original equation we get
$$243 x \equiv 112(\bmod 551)$$

Since $(243,551)=1$, by Lemma 12, we know that $243 x \equiv 112(\bmod 551)$ has integer solutions. Since
$$\begin{array}{l}
551=2 \times 243+65, \quad 243=65 \times 3+48 \\
65=48+17, \quad 48=17 \times 2+14, \quad 17=14 \\
+3, \quad 14=3 \times 4+2, \quad 3=2+1
\end{array}$$

We get
$$\begin{array}{l} 
1=3-2=3-(14-3 \times 4)=5 \times 3-14 \\
=5 \times(17-14)-14=5 \times 17-6 \times 14 \\
=5 \times 17-6 \times(48-17 \times 2)=17 \times 17 \\
-6 \times 48=17 \times(65-48)-6 \times 48 \\
= 17 \times 65-23 \times 48=17 \times 65-23 \\
\times(243-65 \times 3)=-23 \times 243+86 \\
\times 65=-23 \times 243+86 \times(551-2 \\
\times 243)=36 \times 551-195 \times 243
\end{array}$$

So we have
$$-195 \times 243 \equiv 1 \quad(\bmod 551)$$

By equation (35) and Lemma 5, we have
$$-195 \times 243 x \equiv-195 \times 112(\bmod 551)$$

Since $-195 \times 112=200-551 \times 40$, we get
$$-195 \times 112=200(\bmod 551)$$

From equations (36) to (38), we get $x \equiv 200(\bmod 551)$. Since $2755=551 \times 5$ and $200+551=751, 200+551 \times 2=1302, 200+551 \times 3=1853, 200+551 \times 4=2404$, the 5 different solutions to $1215 x \equiv 560 (\bmod 2755)$ are
$$x \equiv 200,751,1302,1853,2404 \quad(\bmod 2755)$$