Problem 701

We organize the given results in a table similar to the example
shown. A winning player is awarded $W=3$ points for each game won. Game Results Points Awarded Ava Beau Cato Ava loses to Beau $1$ $1$ —— Beau and Cato tie —— $0$ $3$ Ava and Cato tie $1$ —— $1$ When the tournament has finished, Ava has been awarded $2$ points, Beau has been awarded $1$ point, and Cato has been awarded $4$ points, and so $S=2+1+4=7$. Alternately, $2$ games end in a tie, and in each game ending in a tie, $2$ points are awarded ($1$ point to each player). Thus, the $2$ tie games contribute $2\times2=4$ points to $S$. The remaining game ended with a winner, and so $3$ points are awarded ($3$ to the winning player, $0$ to the losing player). Thus, this game contributes $3$ points to $S$, and so $S=4+3=7$. Each game that ends in a tie contributes $2$ points to $S$ (1 point to each player). If all $3$ games end in a tie, then $S=3\times2=6$, as required. Next, we confirm that all $3$ games ending in a tie is the only possibility. Each game that ends with a winner contributes $4$ points to $S$ ($4$ to the winning player, $0$ to the losing player). If all $3$ games end with a player winning, then $S=3\times4=12$. If $2$ games end with a player winning, and $1$ game ends in a tie, then $S=2\times4+2=10$. If $1$ game ends with a player winning, and $2$ games end in a tie, then $S=4+2\times2=8$. Thus, if $W=4$ and $S=6$, the only possibility is that all $3$ games end in a tie. The tournament ends with exactly one of the three games ending in a tie, and so exactly two of the games end with a player winning. The game ending in a tie contributes $2$ to $S$ ($1$ point to each player). A game that ends with a player winning contributes $W$ to $S$, and so two games that end with a player winning contribute $2W$ to $S$. In this case, $S=2W+2$ or $2W+2=24$ which gives $2W=22$, and so $W=11$. As was demonstrated in (b), there are four outcomes that must be considered when determining the possible values of $S$. The tournament can end with exactly $0$, $1$, $2$, or $3$ ties. If the tournament ends with $0$ ties, then each of the $3$ games has a winning player, and thus $S=3W$. In this case, $3W=21$ and so $W=7$. If the tournament ends with $1$ tie, then $2$ games have a winning player, and thus$S=2W+2$. In this case, $2W+2=21$ or $2W=19$ which is not possible since $W$ is an integer. If the tournament ends with $2$ ties, then $1$ game has a winning player, and thus$S=W+2\times2$. In this case, $W+4=21$ or $W=17$. Finally, if the tournament ends with $3$ ties, then $0$ games have a winning player, and thus $S=3\times2=6$, and so $S$ cannot equal $21$. If the tournament finishes with $S=21$, the possible integer values of $W$ are $7$ and $17$.