1. How many non-empty subsets of the set $\{0,1, \ldots, 9\}$ have the sum of their elements divisible by three?

(Eliška Macáková)
SOLUTION. We start from the fact that the remainder of the sum of several integers when divided by three is the same as the remainder of the number obtained by summing the remainders of the individual addends.* Therefore, we will divide the 10 given numbers into three sets according to their remainders (when divided by three - we will no longer write this) as follows:

$$
A_{0}=\{0,3,6,9\}, \quad A_{1}=\{1,4,7\}, \quad A_{2}=\{2,5,8\}
$$

Addends from $A_{0}$ do not affect the remainders of the sums being examined. Each of these final remainders is fully determined by two counts: the number of addends from $A_{1}$, which we denote as $x$, and the number of addends from $A_{2}$, which we denote as $y$. Since $x, y \in\{0,1,2,3\}$, by examining all possible pairs $(x, y)$, we can test for which of them the sum of $x$ numbers from $A_{1}$ and $y$ numbers from $A_{2}$ is divisible by three.**

We will now describe (without further explanation) all types of valid selections of numbers from both sets $A_{1}, A_{2}$ and provide their counts.

- In the case where no number is selected from $A_{1}$, we must select 0 or 3 numbers from $A_{2}$. The same conclusion about the selection from $A_{2}$ applies if we select 3 numbers from $A_{1}$. In both cases, we have four options for selecting numbers from $A_{1} \cup A_{2}$ (including one "empty" selection): \{\}$,\{2,5,8\},\{1,4,7\},\{1,4,7,2,5,8\}$.***
- The remaining cases are when we select 1 or 2 numbers from $A_{1}$. Then we must select the same number of numbers from $A_{2}$. Since each three-element set has 3 one-element and 3 two-element subsets, we have a total of $3 \cdot 3 + 3 \cdot 3 = 18$ additional ways to select numbers from $A_{1} \cup A_{2}$ (we will not list them here).

Thus, there are a total of $4 + 18 = 22$ valid selections of numbers from $A_{1} \cup A_{2}$. Each of these can be supplemented with numbers from the 4-element set $A_{0}$ in $2^{4} = 16$ ways (see problem N1), including the "empty" supplementation. The total number of valid selections from $A_{0} \cup A_{1} \cup A_{2}$ is therefore $22 \cdot 16 = 352$, where the "empty" selection is also counted. Therefore, the sought number is 351.

Conclusion. The required property is satisfied by 351 non-empty subsets.