65. Find the remainder when the number

$$
10^{10}+10^{\left(10^{0}\right)}+\ldots+10^{\left(10^{10}\right)}
$$

is divided by 7.
65. First of all, note that \(10^{6}-1=999999\) is divisible by 7 (since \(999999=7 \cdot 142857\)). From this, it easily follows that \(10^{\mathrm{N}}\), where \(N\) is any integer, gives the same remainder when divided by 7 as \(10^{r}\), where \(r\) is the remainder from dividing \(N\) by 6. Indeed, if \(N=6k+r\), then

\[
\begin{aligned}
& 10^{k}-10^{r}=10^{6k+r}-10^{r}=10^{r}\left(10^{6k}-1\right)= \\
& =10^{r} \cdot \left(10^{6}-1\right)\left(10^{6k-6}+10^{6k-12} \ldots+10^{6}+1\right)
\end{aligned}
\]

is divisible by 7.

But any integer power of 10 gives a remainder of 4 when divided by 6; indeed, \(10^{n}-4=\underset{(n-1) \text{ times }}{999 \ldots 9}\) is always divisible by \(2 \cdot 3=6\) (due to the divisibility rules for 2 and 3). Thus, all the complex exponents of the terms in our sum give a remainder of 4 when divided by 6. Consequently, each of these 10 terms gives the same remainder when divided by 7 as the number \(10^{4}\), and the entire sum gives the same remainder as the number

\[
\begin{aligned}
10^{4}+10^{4}+10^{4}+10^{4}+10^{4}+10^{4}+10^{4} & +10^{4}+10^{4}+10^{4}= \\
& =10^{5}=100000=7 \cdot 14285+5
\end{aligned}
\]

Thus, the desired remainder is 5.