1. The average of Pažout's grades is exactly 3. If we did not include three of Pažout's fives in the average, the average of his grades would be exactly 2. Determine the maximum number of ones that Pažout could have received. (Possible grades are 1, 2, 3, 4, 5.)
SOLUTION. Let $s$ be the sum of all Pážout's grades and $p$ their count. We do not know either of these numbers, but we know that

$$
\frac{s}{p}=3 \quad \text { or } \quad s=3 p .
$$

According to the second sentence of the problem, we set up another equation for the unknowns $s, p$ and immediately simplify it:

$$
\frac{s-3 \cdot 5}{p-3}=2 \quad \Leftrightarrow \quad s-15=2 p-6 \quad \Leftrightarrow \quad s=2 p+9
$$

From this, by substituting $s=3 p$, we get the equation $3 p=2 p+9$ with the only solution $p=9$. The sum of the grades is thus equal to $s=3 p=27$.

Besides the three fives, Pážout therefore received another six grades with a sum equal to $27-3 \cdot 5=12$. It could not have been six ones (with a sum of only 6), nor five ones, because the sixth grade would then be $12-5 \cdot 1=7$. However, Pážout could have received four ones, his remaining two grades with a sum of $12-4=8$ would then be either two fours, or a three and a five.

Conclusion. The maximum possible number of Pážout's ones is 4.

For a complete solution, award 6 points. In case of incomplete solutions, evaluate partial steps as follows:

A0. Stating the correct answer without justification: 0 points.

A1. Stating the correct answer along with a suitable example of a set of grades: 1 point.

A2. Translating the verbal problem into mathematical notation (most likely into a system of two equations with two unknowns): 1 point.

A3. Solving the mathematical problem from point A2 or other justification leading to the conclusion that Pážout received 9 grades (or 6 grades, if the three fives are not considered): 4 points.

A4. Justification why Pážout could not have received more than 4 ones: 5 points.

In total, then award $\max (\mathrm{A} 2, \mathrm{~A} 3, \mathrm{~A} 4)+\mathrm{A} 1$ points. In a complete solution, it must be justified that the situation with four ones can actually occur (most likely by an example of a set of grades, as stated in point A1).