pongue

Some students from the seventh and eighth grades of a school participate in a ping pong tournament, where each student plays against all the others exactly once, receiving 1 point for a win and 0 points for a loss. There are ten times as many eighth-grade students as seventh-grade students. The total score of the eighth-grade students is 4.5 times the total score of the seventh-grade students.

a) Verify that if there are $k$ students in the tournament, then the number of games is $\frac{k(k-1)}{2}$.

b) What is the sum of the scores obtained by all the seventh-grade students?
pongue - Solution

a) Each of the $k$ students will play $k-1$ times. Summing the number of games each one plays, we get $k(k-1)$. However, we will have counted each game twice, once for each of the participants in the match. Therefore, the number of games is $\frac{k(k-1)}{2}$.

b) Let $n$ be the number of seventh-grade students and suppose this group of students scored $m$ points. Then the number of eighth-grade students is $10 n$ and the score obtained by them is $4.5 \mathrm{~m}$. Since the total number of participants in the tournament is $n+10 n=11 n$ and each match is worth exactly 1 point, the number of points contested is $\frac{11 n(11 n-1)}{2}$. On the other hand, we know that $m+4.5 m$ corresponds to all the points contested, thus

$$
\begin{aligned}
\frac{11 n(11 n-1)}{2} & =5.5 m \\
n(11 n-1) & =m
\end{aligned}
$$

Since each participant played $11 n-1$ games and each game is worth 1 point, the sum of the points obtained by the seventh-grade students is at most $n(11 n-1)$. Since this number coincides with the score obtained by this group of participants, we can conclude that there are no two seventh-grade students who played against each other, otherwise we must subtract from $n(11 n-1)$ the double counting of such a match. This is only possible if there is only one seventh-grade student. Therefore, $n=1 \mathrm{~and} m=1 \cdot 10=10$.