Write down the first $n$ natural numbers in decimal form on a (fairly long) strip of paper, then cut the strip so that each piece contains only one digit. Put these pieces in a box, mix them up, and draw one at random. Let $p_{n}$ denote the probability that the digit 0 is on the piece of paper drawn. Determine the limit of the sequence $p_{n}(n=1,2, \ldots)$.
P. 158. Let's write the first n natural numbers in decimal form on a (fairly long) piece of paper, then cut the paper into pieces so that each piece contains only one digit. Place these pieces in a box, mix them, and draw one at random. Let $p_{n}$ denote the probability that the digit 0 is on the piece of paper drawn. Determine the limit of the sequence $p_{n}(n=1,2, \ldots)$.

None of the submitted solutions solved this problem. Therefore, the editorial board decided to rephrase the problem and issue it again. This became problem F. 1883. The solution can be found on page 61 of this issue.

Solution. Let $k$ be the number of digits in the decimal representation of $n$. Imagine extending the decimal representation of the numbers less than $n$ to $k$ digits by writing 0s in the places of the missing digits, and write $k$ zeros at the beginning of the sequence. Let $R_{k}$ denote the number of 0s written in this process. After the extension, each written number has $k$ digits, so the total number of digits is $(n+1) k$, and we have

$$
Q_{n}+R_{k}=(n+1) k
$$

During the extension, we wrote a single 0 in the units place, ten 0s in the tens place, and so on, finally writing $10^{k-1}$ 0s in the first column, which has the value $10^{k-1}$. Therefore,

$$
R_{k}=1+10+10^{2}+\ldots+10^{k-1}=\frac{10^{k}-1}{9}
$$

Thus,

$$
0\frac{k}{10} n-2 n=\frac{k-20}{10} \cdot n
$$

According to (2) and (3), similarly, on the one hand,

$$
Q_{n}=(n+1) k-R_{k}>n k-2 n=(k-2) n
$$

and on the other hand,

$$
Q_{n}<(k+1) n
$$

Based on these, we can set bounds for the ratio $\frac{P_{n}}{Q_{n}}$:

$$
\frac{k-20}{k+1} \cdot \frac{1}{10}<\frac{P_{n}}{Q_{n}}<\frac{k+12}{k-2} \cdot \frac{1}{10}
$$

Here, the limit of both sides is $\frac{1}{10}$, since $k$ also tends to infinity as $n$ does. This is indeed the case, as for any large number $A$, there exists an $N$ such that $k \geqq A$ for all natural numbers $n$ greater than $N$. - This completes the proof of the problem statement.