3. In the tournament, each participant was supposed to play exactly one game with each of the remaining participants, but two participants dropped out during the tournament, having played only 4 games each. In the end, the total number of games played turned out to be 62. How many participants were there in total?
Answer: 13.

Solution: Let the total number of participants be $n$. Then, excluding the two who dropped out, the remaining participants played $\frac{(n-2)(n-3)}{2}$ matches. If these two managed to play against each other, then 7 matches were played with their participation, and if they did not, then 8 matches. Thus, we have two cases: $\frac{(n-2)(n-3)}{2}+8=62$ or $\frac{(n-2)(n-3)}{2}+7=62$. Solving these equations, we find that

## MATHEMATICS 10th GRADE

the first equation has no integer solutions, while the second has one positive integer solution $n=13$.

Grading criteria: A correct and justified solution - 7 points. A correct approach but an incorrect answer due to an arithmetic error - 5 points. Noticing the two cases but not completing the second, obtaining the correct answer - $\mathbf{4}$ points. In a trial-and-error solution, not all cases are considered, but the correct answer is obtained - $\mathbf{3}$ points. Only the correct answer with justification that it fits - $\mathbf{2}$ points. Incorrect solution or only the answer - $\mathbf{0}$ points.