5. The number $10^{100^{100^{10}}}$ when divided by 77, the remainder is ( ).
(A) 45
(B) 56
(C) 67
(D) 68
5.C.

Let $x=10^{10^{10^{10}}}$. Note that $77=7 \times 11$, obviously,
$$
x=(11-1)^{10^{10^{10}}} \equiv(-1)^{10^{10^{10}}} \equiv 1(\bmod 11) \text {. }
$$

Since 7 and 10 are coprime, by Fermat's Little Theorem, we have $10^{6} \equiv 1(\bmod 7)$.
Next, find the remainder $r(0 \leqslant r \leqslant 5)$ when $10^{10^{10}}$ is divided by 6. Note that, $6=2 \times 3$, obviously, $r \equiv 0(\bmod 2)$, and
$$
r \equiv 10^{10^{10}} \equiv\left(3^{2}+1\right)^{10^{10}} \equiv 1(\bmod 3) .
$$

Thus, by substituting $r=0,2,4$ one by one, we find that only $r=4$.
Let $10^{10^{10}}=6 k+4\left(k \in Z_{+}\right)$, then
$$
\begin{array}{l}
x \equiv 10^{6 k+4} \equiv\left(10^{6}\right)^{k} \times 10^{4} \\
\equiv 10^{4} \equiv 3^{4} \equiv 9^{2} \equiv 2^{2} \equiv 4(\bmod 7) .
\end{array}
$$

From equation (1), we have $x=11 y+1(y \in Z)$, substituting into equation (2) gives
$$
11 y+1=4(\bmod 7) \text {, }
$$

which is $11 y=3(\bmod 7)$,
or $-3 y \equiv 3(\bmod 7)$.
Thus, $y \equiv-1(\bmod 7)$.
Let $y=7 z+6(z \in Z)$, then
$$
x=11(7 z+6)+1=77 z+67 \text {. }
$$

Therefore, the remainder when $x$ is divided by 77 is 67.