Problem 504

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 $17$.