G1.3 Determine the remainder of the 81 -digit integer $\overline{111 \cdots 1}$ divided by 81 .
$$
\begin{aligned}
\overline{81 \cdots 1} \text { digits } & =10^{80}+10^{79}+\cdots+10+1 \\
& =\left(10^{80}-1\right)+\left(10^{79}-1\right)+\cdots+(10-1)+81 \\
& =\underbrace{\overline{9 \cdots \cdots 9}}_{80 \text { digits }}+\underbrace{\overline{99 \cdots 9}}_{79 \text { digits }}+\cdots+9+81 \\
& =9 \times(\underbrace{\overline{11 \cdots 1}}_{80 \text { digits }}+\underbrace{\overline{11 \cdots 1}}_{79 \text { digits }}+\cdots+1)+81
\end{aligned}
$$

Let $x=\underbrace{\overline{11 \cdots 1}}_{80 \text { digits }}+\underbrace{\overline{11 \cdots 1}}_{79 \text { digits }}+\cdots+1 \equiv 80+79+\cdots+1(\bmod 9)$
$$
x \equiv \frac{81}{2} \cdot 80 \equiv 9 \times 40 \equiv 0 \bmod 9
$$
$x=9 m$ for some integer $m$
$$
\overline{81 \text { digits }} \overline{11 \cdots 1}=9 \times 9 m+81 \equiv 0(\bmod 81)
$$

The remainder $=0$