Problem 1. For each of the nine natural numbers $n, 2 n, 3 n, \ldots, 9 n$, the leftmost digit was written down. Can it be that for some natural $n$, among the nine written digits, there are no more than four different ones?

$(2022-69$ / Tolpygo A.)
Answer: Yes, it can.

Solution. For example, $n=25$ works. For this value of $n$, we get the numbers

$$
25,50,75,100,125,150,175,200,225
$$

Each of these numbers starts with one of the four digits $1,2,5,7$.

Comment. It can be proven that it is impossible to have fewer than four different digits. Exactly four different digits occur for natural numbers in the intervals of the form $[25 \underbrace{0 \ldots 0}_{k-1}, 2 \underbrace{9 \ldots 9}_{k}]$ and $[3 \underbrace{3 \ldots 3}_{k-1} 4,3 \underbrace{9 \ldots 9}_{k}]$, where $k$ is any natural number.