In a mathematics competition, $A, B, C, D, E$ and $F$ participated. After the first test, the organizer said: "3 contestants have 10 points, the others have 7. Everyone should submit their guess about who they think earned 10 points." The contestants submitted the following guesses:

$$
A, B, D ; \quad A, C, E ; \quad A, D, E ; \quad B, C, E ; \quad B, D, E ; \quad C, D, E
$$

The organizer announced that no one guessed all three qualifiers correctly, 3 guesses got 2 right, 2 guesses got 1 right, and 1 guess got none right.

Based on this, who had 10 points?
I. solution. The tips contained a total of $3 \cdot 2 + 2 \cdot 1 = 8$ correct hits. The number of tips given for each contestant is respectively: $3,3,3,4,5,0$. These numbers can only add up to 8 in the form $8 = 5 + 3 + 0$. Therefore, the contestants who received 5, 3, and 0 tips could have 10 points, with 5 and 0 tips being given only to $E$ and $F$, respectively, so they have 10 points. The only one who could still have 10 points is $A$, $B$, or $C$. Except for the first tip, all tips include $E$, so they must have at least one correct hit. Therefore, only the first tip could be a 0-hit tip, so neither $A$ nor $B$ could have advanced, leaving $C$ with 10 points.

If $C$, $E$, and $F$ have 10 points, then the tips respectively hit $0,2,1,2,1,2$ correct contestants, which is the solution to our problem.

Miklós Sliz (Tatabánya, Ságvári E. Secondary School, 8th grade)

II. solution. Each tip selects 3 out of the first 5 contestants. If all three 10-pointers were among the first five, then each tip would have at least 1 hit, which is not possible. Therefore, the sixth contestant, $F$, has 10 points.

The three 2-hit tips only hit the other two 10-pointers. Of these three tips, at least two are among the last five tips, and $E$ appears on all of them. If $E$ were the one who incorrectly appears on them, then these two 2-hit tips would be identical. Since all tips are different, $E$ must have 10 points, and each of the last five tips has at least one hit. According to this, only the first tip can be a 0-hit tip, and $C$ has the remaining 10 points.

$$
\text{Béla Kartaly (Szolnok, Verseghy F. Gymnasium, 1st grade)}
$$

Note. We can choose 3 out of 6 contestants in 20 different ways. If we determine the number of hits for each of these combinations based on the given tips, we can select the correct one. It is not worth examining the combinations that appear in the tips, as there would be a 3-hit tip next to them. Out of the remaining 14 combinations, it is enough to examine only the six that are the complements (opposites) of the given tips (e.g., the complement of $A, B, C$ is $D, E, F$), since we know there is a 0-hit tip. By this method, with a bit more mechanical work, we can also arrive at the solution.

Zoltán Dávid (Pécs, Széchenyi I. Gymnasium, 1st grade)