## Task B-1.6.

The sum of the fractions $\frac{12}{23}+\frac{1212}{2323}+\frac{121212}{232323}+\cdots+\frac{1212 \ldots 12}{2323 \ldots 23}$ is 528. The number of digits 1 and 2 in the numerator and the number of digits 2 and 3 in the denominator of these fractions increase by one. How many times does the digit 2 appear in the last fraction?
## Solution.

Starting from the second addend, all fractions can be simplified as follows:

$$
\begin{array}{rlrl}
\frac{1212}{2323} & =\frac{12 \cdot 100+12}{23 \cdot 100+23}=\frac{12 \cdot 101}{23 \cdot 101}=\frac{12}{23} & 2 \text { points } \\
\frac{121212}{232323} & =\frac{12 \cdot 10^{4}+12 \cdot 10^{2}+12}{23 \cdot 10^{4}+23 \cdot 10^{2}+23}=\frac{12 \cdot 10101}{23 \cdot 10101}=\frac{12}{23} & & 2 \text { points } \\
\vdots & & 2 \text { points } \\
\frac{1212 \ldots 12}{2323 \ldots 23} & =\frac{12 \cdot 101 \ldots 01}{23 \cdot 101 \ldots 01}=\frac{12}{23} &
\end{array}
$$

If the number of addends in the given sum is denoted by $n$, then $n \cdot \frac{12}{23}=528$, from which we get that the number of addends is $n=1012$.

In the first fraction, the digit 2 appears 2 times, in the second 4 times, in the third 6 times, ... in the $n$-th $2 n$ times.

Thus, in the last fraction, the digit 2 appears $2 \cdot 1012=2024$ times.

Note: If a student writes without any argumentation that each of the fractions in the given sum is equal to $\frac{12}{23}$, the student should be awarded 2 points out of the 6 points allocated for that part of the task.