$105$ games were played in a chess tournament, every player played with every other player, how many players participated? $105$ total games
$12$, $13$, $14$, $15$ or $53$ players?
 A: Let $n$ be the number of players. There was a game for every pair of players, so there must be $105$ pairs of players. An $n$-element set has $\binom{n}2=\frac{n(n-1)}2$ $2$-element subsets, so 
$$\frac{n(n-1)}2=105\;.$$
You should be able to solve that for $n$ with little difficulty. Or you could simply test each of the choices to see which one satisfies the equation.
A: Hint: if there are $n$ players, each will play $n-1$ games, but each game will be counted twice.
A: Well, I don't know about chess, but sometimes it depends on the game you're playing. For example in a football(soccer) match, because we have home-away system, sometimes it's different that $A$ plays against $B$ or $B$ plays against $A$. In that case, the order matters, and if $m$ games are played, then the number of teams participating in the tournament is $n(n-1)=m$ because we can choose $n$ teams at first, and then we have to choose from the $n-1$ teams left, because a team can not play with itself. But if there is no home-away system, then each match is counted twice because the order doesn't matter and $A$ plays against $B$ is the same as $B$ plays against $A$, so the answer is $\displaystyle \frac{n(n-1)}{2}=m$ which is the same as the number of ways you can choose $2$ teams/players from $n$ teams/players to play against each other.
I believe chess doesn't have home-away matches and two individuals play against each other only once, so it must be $\displaystyle \frac{n(n-1)}{2}=105$:
$$ \frac{n(n-1)}{2}=105 \implies n(n-1)=210 \implies n^2-n-210=0$$
$$ \implies (n-15)(n+14)=0 \implies n=15.$$
