## Task B-2.7.

Students of a certain school participated in a table tennis tournament. The rules stipulate that each pair of competitors must play exactly one match. During the tournament, students $A, B$, and $C$ withdrew from further competition. Student $A$ withdrew after playing two matches, $B$ after three, and $C$ after five matches. The other competitors completed the tournament. If a total of 197 matches were played during the tournament, how many students participated in the tournament?
## Solution.

Let $n$ be the number of students who participated in the tournament. After three students dropped out during the competition, each of the $n-3$ students played exactly one game with each of the remaining students until the end of the tournament.

Thus, among the $n-3$ students, $\frac{(n-3)(n-4)}{2}$ games were played.

2 points

To this number, we need to add the number of games $(p)$ in which students $A$, $B$, and $C$ participated before they dropped out. There could be at most $2+3+5=10$ such games, assuming $A$, $B$, and $C$ did not play against each other.

Since they could have played up to $\frac{3 \cdot 2}{2}=3$ games against each other, the number $p$ can be 10, 9, 8, or 7, so we have $\frac{(n-3)(n-4)}{2}+p=197, p \in\{10,9,8,7\}$.

If $p=10$, from $\frac{(n-3)(n-4)}{2}+10=197$ we get the equation $(n-3)(n-4)=374$, which has no solutions in the set of natural numbers because the number $374=2 \cdot 11 \cdot 17$ cannot be written as the product of two consecutive natural numbers.

If $p=9$, from $\frac{(n-3)(n-4)}{2}+9=197$ we get the equation $(n-3)(n-4)=376$, which has no solutions in the set of natural numbers because the number $376=2^{3} \cdot 47$ cannot be written as the product of two consecutive natural numbers.

If $p=8$, from $\frac{(n-3)(n-4)}{2}+8=197$ we get the equation $(n-3)(n-4)=378$, which also has no solutions in the set of natural numbers because the number $378=2 \cdot 3^{3} \cdot 7$ cannot be written as the product of two consecutive natural numbers.

If $p=7$, from $\frac{(n-3)(n-4)}{2}+7=197$ we get the equation $(n-3)(n-4)=380$. Since $380=19 \cdot 20$, we conclude that $n-4=19$, i.e., $n=23$.

Therefore, 23 students participated in the competition.

Note: In each of the described cases, students could have reached the same conclusions by solving the quadratic equation or in some other way, so the procedure should be graded with the expected number of points.

## COUNTY COMPETITION IN MATHEMATICS

## 3rd grade - high school - B variant

March 29, 2021.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.