2. How many positive values can the expression

$$
a_{0}+3 a_{1}+3^{2} a_{2}+3^{3} a_{3}+3^{4} a_{4}
$$

take if the numbers $a_{0}, a_{1}, a_{2}, a_{3}$ and $a_{4}$ are from the set $\{-1,0,1\}$?
III/2. First, we observe that the numbers $a_{0}+3 a_{1}+3^{2} a_{2}+3^{3} a_{3}+3^{4} a_{4}$ and $a_{0}^{\prime}+3 a_{1}^{\prime}+3^{2} a_{2}^{\prime}+3^{3} a_{3}^{\prime}+3^{4} a_{4}^{\prime}, a_{i}, a_{i}^{\prime} \in\{-1,0,1\}$, are equal if and only if $a_{i}=a_{i}^{\prime}$ for $i=0,1,2,3,4$.

If $a_{4}=1$, the values of $a_{0}, \ldots, a_{3}$ can be arbitrary (we get $3^{4}$ different numbers). If $a_{4}=-1$, the number will be negative regardless of the values of the other coefficients. If $a_{4}=0$, we repeat a similar reasoning for $a_{3}$. Therefore, the total number of possible numbers is $3^{0}+3^{1}+3^{2}+3^{3}+3^{4}=$ $1+3+9+27+81=121$.

Observation that the non-zero coefficient with the highest index must be positive: 2 points. Calculation of the number of possibilities for this case: 2 points. Sum $3^{0}+3^{1}+3^{2}+3^{3}+3^{4}=121$: 3 points.

If the contestant correctly writes all 121 possibilities: 7 points. If the contestant (incorrectly!) concludes that this is just a slightly modified representation in the ternary system and there are therefore $3^{5}-1$ possibilities: 0 points. If the contestant (correctly!) concludes that this is just a slightly modified representation in the ternary system and we are only interested in those numbers that are greater than $\overline{11111}_{(3)}=121$, but not greater than $\overline{22222}_{(3)}=242$: 7 points.