5. Let $p$ be a prime number greater than 3, then the remainder when $p^{2}$ is divided by 6 is ( ).
(A) 5
(B) 3
(C) 1
(D) Cannot be determined, depends on the value of $p$
5.C.

For primes greater than 3, when divided by 3, the remainder is 1 or 2, which can be expressed as $3m+1$ or $3m+2$. Since primes greater than 3 are necessarily odd, when $p=3m+1$, $m$ must be even $(m=2k)$; when $p=3m+2$, $m$ must be odd $(m=2k-1)$.
Therefore, $p$ can be expressed as $p=6k \pm 1$, yielding
$$
p^{2}=36k^{2} \pm 12k+1=12k(3k \pm 1)+1 \text{. }
$$

Thus, $p^{2} \equiv 1(\bmod 6)$.
Note: The proof has already shown that $p^{2} \equiv 1(\bmod 12)$.