$2 \cdot 48$ Try to find the remainder when $10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$ is divided by 7.
(5th Moscow Mathematical Olympiad, 1939)
[Solution] Let $A=10^{10}+10^{10^{2}}+10^{10^{3}}+\cdots+10^{10^{10}}$.
First, we prove that if $6 \mid n-r$, then $7 \mid 10^{n}-10^{r},(n>r)$.
In fact,
$$
\begin{aligned}
10^{n}-10^{r} & =10^{r}\left(10^{n-r}-1\right)=10^{r}\left(10^{6 k}-1\right) \\
& =\left(\left(10^{6}\right)^{k}-1\right) \cdot 10^{r} .
\end{aligned}
$$

Since $10^{6} \equiv 1(\bmod 7)$, we have
$$
\left(10^{6}\right)^{k}-1 \equiv 0 \quad(\bmod 7) .
$$

Thus,
$$
7 \mid 10^{n}-10^{r} \text {. }
$$

On the other hand, if $6110^{k}-10 \quad(k \geqslant 1)$, then
$$
\begin{array}{l}
A-10 \cdot 10^{10}+10 \cdot 10^{10} \\
=\left(10^{10}-10^{10}\right)+\left(10^{10^{2}}-10^{10}\right)+\cdots+\left(10^{10^{10}}-10^{10}\right)+10 \cdot 10^{10} \text {. } \\
7 \mid 10^{10^{k}}-10^{10} \text {. } \\
\end{array}
$$

Since $6+10^{k}-10$, then

Thus we have
$$
\begin{aligned}
A & \equiv 10 \cdot 10^{10} \quad(\bmod 7) \\
& =10^{11} \\
& =(7+3)^{11} \\
& \equiv 3^{11} \\
& \equiv 3^{5} \cdot 3^{6} . \quad(\bmod 7)
\end{aligned}
$$

Since
$$
\begin{array}{ll}
3^{5} \equiv 5 & (\bmod 7) \\
3^{6} \equiv 1 & (\bmod 7) \\
A \equiv 5 & (\bmod 7)
\end{array}
$$

Thus,
the remainder when $A$ is divided by 7 is 5.