Maths Olympiad Prep

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Combinatorics Difficulty 3.1 AMC 10/12 Prove it Canada

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 2727 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 NameGames PlayedNumber of WinsNumber of LossesNumber of TiesTotal Points
PP27101423
QQ27
RR2725
SS27

How many ties did Team PP have at the end of the season?
Team QQ had 22 more wins than Team PP and 44 fewer losses than Team PP. How many total points did Team QQ have at the end of the season?
Explain why Team RR could not have finished the season with exactly 66 ties.
At the end of the season, Team SS had 44 more wins than losses. Show that Team SS must have finished the season with a total of 3131 points.

Solution

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

Thus, Team PP had 271014=327-10-14=3 ties at the end of the season.
Team QQ had 2 more wins than Team PP, or 10+2=1210+2=12 wins.

Team QQ had 4 fewer losses than Team PP, or 144=1014-4=10 losses.

Since Team QQ played 27 games, they had 271210=527-12-10=5 ties.

At the end of the season, Team QQ had a total of (2×12)+(0×10)+(1×5)(2\times12)+(0\times10)+(1\times5) or 29 points.
Solution 1
Assume that Team RR finished the season with exactly 6 ties.

Since 6 ties contribute 6 points to their points total, then Team RR earned the remaining 256=1925-6=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 RR could not have finished the season with exactly 6 ties.

Solution 2

Assume that Team RR finished the season with exactly ww wins.

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

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

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

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

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

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

Since Team SS played 27 games, then each of their remaining 27(+4)=23227-\ell-(\ell+4)=23-2\ell games resulted in a tie.

Therefore, Team SS finished the season with a total of (2×(+4))+(0×)+(1×(232))(2\times(\ell+4)) + (0\times \ell) + (1\times (23-2\ell)) or 2+8+232=312\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 4×272=54\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 2×54=1082\times54=108.

The table shows that Team PP finished with 23 points, Team RR had 25 points, and in part (b) we determined that Team QQ had 29 points at the end of the season.

Therefore, Team SS finished the season with 108232529=31108-23-25-29=31 points.

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