7. Use the previous problem to solve (i) $3 x \equiv 1(\bmod 125)$; (ii) $5 x \equiv 1(\bmod 243)$.
7. (i) $3 x \equiv 1\left(\bmod 5^{3}\right), 3 \cdot 2 \equiv 1(\bmod 5), 1-(1-3 \cdot 2)^{3}=126$, so $x=42$ is a solution to the original congruence equation. (ii) $5 x \equiv 1\left(\bmod 3^{5}\right) .5 \cdot(-1) \equiv 1(\bmod 3) .1-(1-5 \cdot(-1))^{5}=$ $-7775 . x=-1555$ is a solution.