106. Several chess players participated in a match tournament, where each participant played several games against every other participant. How many rounds did this competition consist of, if a total of 224 games were played?
Solution. Let $x$ be the number of participants in the competition, and $y$ be the number of rounds. In one round, each chess player plays $(x-1)$ games, and all participants together play $\frac{x(x-1)}{2}$ games. Therefore, the total number of games played in the competition is $\frac{x(x-1) y}{2}$. Consequently, $x(x-1) y=448$. Listing the divisors of 448, we notice that only two of them (7 and 8) differ by 1. From the obtained equation, it is easy to find $x=8, y=8$.

Answer: The tournament was held in 8 rounds.