4. (3 points) Does there exist a five-digit natural number with the sum of its digits equal to 31, which is divisible by 31? Answer: Yes, for example $93775=31 \cdot 3025$.
In general, the solution is contained in the answer, but let's explain how to find such a number. The sum of the digits of the product gives the same remainder when divided by 9 as the sum of the original numbers. Let's represent 23 as the difference between a number divisible by 5 and a number divisible by 9: $31=5 \cdot 8-9$. Therefore, we need to find a three-digit multiplier with a sum of digits equal to 10.

There are several dozen suitable numbers.