## Task 4 - 050714

Rolf presents his friend with the following problem:

In a chess tournament, everyone played exactly one game against each other. In total, 28 games were played.

How many participants were there in this tournament?
Since everyone plays against everyone else exactly once, if $\mathrm{n}$ is the number of participants, then: $\mathrm{n}(\mathrm{n}-1) / 2 = 28$, so $\mathrm{n}(\mathrm{n}-1) = 56$. This equation is satisfied by the natural number 8. Since for any natural number n, larger values of n always result in larger values of $n(n-1)$, there is no other natural number that satisfies this equation. Therefore, 8 people participated in the chess tournament.

Solutions of the I. Round 1965 adopted from [5]

### 4.7.2 II. Round 1965, Class 7