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Problem 496

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Prove it CEMC Galois · Canada · 2021

The organizer for a sports league with four teams has entered some of the end-of-season data into the table shown. Each team played $27$ games and each game resulted in a win for one team and a loss for the other team, or in a tie for both teams. Each team earned 2 points for a win, 0 points for a loss, and 1 point for a tie.

Team Name
Games Played
Number of Wins
Number of Losses
Number of Ties
Total Points

$P$
27
10
14

$Q$
27

$R$
27

$S$
27

How many ties did Team $P$ have at the end of the season?
Team $Q$ had $2$ more wins than Team $P$ and $4$ fewer losses than Team $P$ . How many total points did Team $Q$ have at the end of the season?
Explain why Team $R$ could not have finished the season with exactly $6$ ties.
At the end of the season, Team $S$ had $4$ more wins than losses. Show that Team $S$ must have finished the season with a total of $31$ points.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Team $P$ played 27 games which included 10 wins and 14 losses.

Thus, Team $P$ had $27-10-14=3$ ties at the end of the season.
Team $Q$ had 2 more wins than Team $P$ , or $10+2=12$ wins.

Team $Q$ had 4 fewer losses than Team $P$ , or $14-4=10$ losses.

Since Team $Q$ played 27 games, they had $27-12-10=5$ ties.

At the end of the season, Team $Q$ had a total of $(2\times12)+(0\times10)+(1\times5)$ or 29 points.
Solution 1
Assume that Team $R$ finished the season with exactly 6 ties.

Since 6 ties contribute 6 points to their points total, then Team $R$ earned the remaining $25-6=19$ points as a result of their wins.

However, each win contributes 2 points to the total, and thus it is not possible to earn an odd number of points from wins.

Therefore, Team $R$ could not have finished the season with exactly 6 ties.

Solution 2

Assume that Team $R$ finished the season with exactly $w$ wins.

If Team $R$ finished with exactly 6 ties, then they finished the season with a total of $(2\times w)+(1\times6)$ or $2w+6=2(w+3)$ points (they earn 0 points for losses).

Since $w$ is an integer, then $w+3$ is an integer and so $2(w+3)$ is an even integer.

However, this is not possible since Team $R$ finished the season with 25 points, an odd number of points.

Therefore, Team $R$ could not have finished the season with exactly 6 ties.
Solution 1

Let the number of losses that Team $S$ had at the end of the season be $\ell$ .

Team $S$ had 4 more wins than losses and thus finished the season with $\ell+4$ wins.

Since Team $S$ played 27 games, then each of their remaining $27-\ell-(\ell+4)=23-2\ell$ games resulted in a tie.

Therefore, Team $S$ finished the season with a total of $(2\times(\ell+4)) + (0\times \ell) + (1\times (23-2\ell))$ or $2\ell+8+23-2\ell=31$ points.

Solution 2

Each of the 4 teams played 27 games, 2 teams played in each game, and so the season finished with a total of $\frac{4\times27}{2}=54$ games played.

Each of the 54 games resulted in a total of 2 points being awarded (either 2 points to a winning team and 0 to the losing team or 1 point to each of the two teams that tied).

Thus, the total points earned by all 4 teams at the end of the season was $2\times54=108$ .

The table shows that Team $P$ finished with 23 points, Team $R$ had 25 points, and in part (b) we determined that Team $Q$ had 29 points at the end of the season.

Therefore, Team $S$ finished the season with $108-23-25-29=31$ points.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.