Let's determine the 73rd digit from the end of $(\underbrace{111 \ldots 1}_{112 \text{ digits}})^{2}$.
I. solution. If we perform the multiplication, then each of the partial products consists entirely of ones. If we add these, then on the last place there is $1 \mathrm{db}$, before that $2 \mathrm{db}, \ldots$, and at the 112th place from the right, 112 ones must be added. Thus, proceeding from the back, the digits of the product are $1,2,3, \ldots, 9$. At the tenth place (where the sum is 10), the digit will be 0, and the carry is 1. Thus, the next digit is not 1, but 2, and then it increases by one again. After a 0, 2 always follows, since the carry increases by one. Thus, at every 9th place value, there is a 9, so at the 72nd place as well. Therefore, the 73rd digit from the back will be 0, as it follows the 9 in the previous sequence.

Ruppert László (Pécs, Janus Pannonius Gymnasium, 11th grade)

II. solution. The 73rd digit from the back of $\underbrace{111 \ldots 1^{2}}_{112 \mathrm{db}}$ is the same as the 73rd digit from the back of

$$
A=\underbrace{111 \ldots 1}_{73 \mathrm{db}}+\underbrace{111 \ldots 10}_{72 \mathrm{db}}+\underbrace{111 \ldots 1}_{71 \mathrm{db}} 00+\cdots+11 \underbrace{000 \ldots 0}_{71 \mathrm{db}}+1 \underbrace{000 \ldots 0}_{72 \mathrm{db}}
$$

This can also be written in the following form:

$$
\begin{aligned}
A= & \frac{10^{73}-1}{9}+10 \cdot \frac{10^{72}-1}{9}+10^{2} \cdot \frac{10^{71}-1}{9}+\cdots+ \\
& +10^{71} \cdot \frac{10^{2}-1}{9}+10^{72} \cdot \frac{10-1}{9}= \\
= & 73 \cdot \frac{10^{73}}{9}-\frac{1+10+\cdots+10^{72}}{9}= \\
= & 8 \cdot 10^{73}+\frac{10^{73}-1}{9}-\frac{1+10+\cdots+10^{72}}{9}+\frac{1}{9}= \\
= & 8 \cdot 10^{73}+\frac{8}{9} \cdot \underbrace{111 \ldots 1}_{73 \mathrm{db}}+\frac{1}{9}=8 \cdot 10^{73}+\frac{1}{9} \cdot \underbrace{888 \ldots 8}_{72 \mathrm{db}} 9 .
\end{aligned}
$$

Since $\frac{1}{9} \cdot \underbrace{888 \ldots 8}_{72 \text { db }} 9$ is an integer and consists of fewer than 73 digits, and $8 \cdot 10^{73}$ ends in 73 zeros, the 73rd place from the back in their sum is 0.

Based on this, the answer to the question is: The 73rd digit from the back of $\underbrace{111 \ldots 1^{2}}_{112 \mathrm{db}}$ is 0.