9. (i) Determine whether $121^{6}-1$ is divisible by 21.
(ii) Find the remainder when $8^{4965}$ is divided by 13.
(iii) Let $p$ be any prime number other than 2 and 5. Prove that: $p \mid \underbrace{99 \cdots 9}_{(p-1) k \uparrow}, k$ is any positive integer.
9. (i) Solution: Since $\varphi(21)=(3-1)(7-1)=12$, and $121 = 11^{12}$, and $(11,21)=1$. By Theorem 1, $11^{12} \equiv 1(\bmod 21)$, so $21 \mid\left(121^{6}-1\right)$.
(ii) Solution: Since $\varphi(13)=12$, and $4965=413 \times 12+9$. By Theorem 2, $8^{12} \equiv 1(\bmod 13)$, so $8^{4955} \equiv 8^{9}(\bmod 13)$. Also, $8^{2} \equiv -1(\bmod 13)$, so $8^{9}=8^{8} \cdot 8 \equiv 8(\bmod 13)$. Therefore,
$$8^{4965} \equiv 8(\bmod 13)$$
(iii) Proof: Since $p \neq 2,5$, we have $(10, p)=1$. Thus, $\left(10^{k}, p\right)=1$. By Theorem 2, $\left(10^{k}\right)^{p-1} \equiv 1(\bmod p)$, and
$$\left(10^{k}\right)^{p-1}-1=\underbrace{99 \cdots 9}_{(p-1) k \uparrow},$$

so
$$p \mid \underbrace{99 \cdots 9}_{(p-1) k \uparrow} .$$