5. Can you use the previous question to propose a method for solving a linear congruence equation? Use the method you propose to solve (i) $6 x \equiv 7(\bmod 23)$; (ii) $5 x \equiv 1(\bmod 12)$. Point out what to pay attention to when applying this method.
5. (i) $m=23, a=6$. $[23 / 6]=3,(3,23)=1$. So the original equation is equivalent to $5 x \equiv -21 \equiv 2(\bmod 23) .[23 / 5]=4,(4,23)=1, 3 x \equiv -8(\bmod 23) .[23 / 3]=7,(7,23)=1$, $2 x \equiv 10(\bmod 23), x \equiv 5(\bmod 23)$. (ii) $[12 / 5]=2,(2,12)=2$, so this method should not be used.