$\underline{\text { Gopovanov A.S. }}$The sum of the digits in the decimal representation of a natural number $n$ is 100, and the sum of the digits of the number $44 n$ is 800. What is the sum of the digits of the number $3 n$?
Note that $44 n$ is the sum of 4 instances of the number $n$ and 4 instances of the number $10 n$.

If we add these numbers digit by digit, then in each digit place, there will be the sum of the quadrupled digit from the same place in the number $n$ and the quadrupled digit from the next place. If no carries occur in the process, then each digit of the number $n$ is added 8 times, and the sum of the digits in all places is 800. With carries, the sum of the digits obviously decreases (since 10 is subtracted from one place and only 1 is added to another). Therefore, in the situation described in the problem, no carries occur. This means, in particular, that any digit of the number $n$ does not exceed 2. Then, when multiplying $n$ by 3, each of its digits is simply multiplied by 3, and thus the sum of the digits.

Therefore, the sum of the digits of the number 3 is 300.

## Problem