Problem 5. Vasya wrote a correct statement:

"In this phrase, $1 / 3$ of all digits are the digit 3, and $1 / 2$ of all digits are the digit 1."

And Kolya wrote the phrase:

"In this phrase, $1 / \ldots$ of all digits are the digit *, the fractions of digits * and * are equal and are $1 / \ldots$, and the fraction of all other digits is $1 / \ldots$."

Insert three different digits in place of the asterisks, and three different numbers in place of the ellipses so that the statement is correct.

$[6$ points] (A. V. Shapovalov)
Answer. "In this phrase, $1 / 2$ of all digits are the digit 1, the shares of digits 2 and 5 are the same and equal to $1 / 5$, and the share of all other digits is 1/10" (or "…the shares of digits 0 and 2 are the same…" or "…the shares of digits 0 and 5 are the same…").

Solution. Let's explain how the problem could be solved (this was not required from the participants).

If any of the numbers replaced by ellipses is at least three digits long, then there are at least 100 digits in this phrase, which is impossible (then one of the numbers consists of at least 30 characters, but then the total number of digits is not less than $10^{30}$, and so on - it is clear that this cannot be). Therefore, all numbers are either single-digit or double-digit, and the total number of digits is between 9 and 12.

The digit "1" is not less than 4, and its share is not less than $4 / 12 = 1 / 3$, so the denominator of the first fraction is a single-digit number, i.e., there are fewer than 12 digits.

All denominators are divisors of the number of digits, greater than one. Therefore, there cannot be 11 digits (the number 11 has no single-digit divisors other than one), nor 9 digits (four terms of the form $1 / 3$ and $1 / 9$ do not sum to 1).

Thus, there are 10 digits in total, and the share of the digit "1" is $1 / 2$. The other fractions can be equal to $1 / 5$ and $1 / 10$. The sum of the shares must be equal to $1: 1 / 2 + 1 / 5 + 1 / 5 + 1 / 10 = 1$. In the denominators, the digits 0, 2, and 5 appear once each, and any two can be mentioned explicitly, then their share will be $1 / 5$, and the share of the only remaining digit will be $1 / 10$.