1. Find all natural numbers for which the following is true: if we add the sum of its digits to the number, we get 313.
I/1. The numbers we are looking for are trimestral. We write: $n=100 a+10 b+c$. Then it holds that 100a+10b+c+a+b+c=313, or $101 a+11 b+2 c=313$. We need to think about 2 possible values for the digit $a$, namely 2 and 3.

First, let's choose $a=2$. Then $11 b+2 c=111$, which is only possible if we choose $b=9$ and $c=6$. One of the numbers we are looking for is 296.

Now let's consider the possibility $a=3$. Then $11 b+2 c=10$, which means $b=0$ and $c=5$. The other number with this property is 305.

The numbers are trimestral: $\mathbf{2}$ points. The equation $100 a+10 b+c+a+b+c=313$ (or its equivalent $101 a+11 b+2 c=313$): $\mathbf{1}$ point. Analysis of the case $a=2: \mathbf{1}$ point. The number sought is 296: 1 point. Analysis of the case $a=3: 1$ point. The number sought is 305: 1 point.

If the contestant provides both solutions but does not prove that they are the only ones: $1+1$ points. If the contestant systematically checks different possibilities and verifies ALL possibilities: 7 points. If the contestant does not verify all possibilities: 1 point for each correct solution.