2023 AMC 10A Problem 16

P16-20~4 minPrint

Games & Processes Divisibility & Factors Ratios, Percents & Averages Use the answer choices (mod checks, size, form) to eliminate or select Bound the quantity above/below or estimate to pin it down

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

Show solution without answering

Key insight

Every game has exactly one winner, so total wins equal total games. Left wins are 1.4 times right wins, which makes the total a multiple of 12.

Solution

Count the players. Say there are $L$ left-handers. Then there are $2L$ right-handers, so $3L$ players in all. The number of players is a multiple of $3$ .
Count the games. Every pair of players meets once. With $n$ players, each plays $n - 1$ others, and each game gets counted by both players, so there are $\frac{n(n-1)}{2}$ games. Try $n = 3, 6, 9, 12$ :

players	3	6	9	12
games	3	15	36	66

Of these, $15$ , $36$ and $66$ are among the answer choices.
3. Use the wins. Each game has exactly one winner, so total wins = total games. Suppose right-handers won $5$ games. Then left-handers won $40\%$ more, which is $7$ . The wins come in the ratio $5 : 7$ , so the total number of wins is a multiple of $5 + 7 = 12$ .
4. Match the two lists. Among $15$ , $36$ , $66$ , only $36$ is a multiple of $12$ .
5. Check it can really happen. With $9$ players ( $3$ left, $6$ right) and $36$ games, right-handers win $15$ and left-handers win $21$ . The left-handers play $3$ games among themselves and $3 \times 6 = 18$ games against right-handers. If they win all $18$ of those, they have $3 + 18 = 21$ wins, and the right-handers have the $15$ games among themselves. It works.

The answer is $\boxed{\textbf{(B)}\ 36}$ .

Why this works

Two separate facts each limit the total. The number of games must come from a whole number of players, and the win ratio $5 : 7$ forces a multiple of $12$ . Only one answer choice passes both tests. In any tournament problem, remember that total wins equal total games.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let there be $L$ left-handed and $2L$ right-handed players, $3L$ in all. Every pair plays once, so the number of games is
$\binom{3L}{2} = \frac{3L(3L-1)}{2}.$
Trying $L = 1, 2, 3, 4$ gives $3, 15, 36, 66$ games; only $15, 36, 66$ appear among the choices.

Each game produces exactly one win. If right-handers won $R$ games, left-handers won $1.4R$ , so the total is $2.4R = \frac{12R}{5}$ . For this to be an integer $R$ must be a multiple of $5$ , and then the total is a multiple of $12$ . Of $15, 36, 66$ , only $36$ is divisible by $12$ .

Sanity check for $L = 3$ : nine players, $36$ games, left-handers win $21$ and right-handers win $15$ . The three left-handers play $3$ games among themselves and $18$ against right-handers; winning all $18$ plus their $3$ internal games gives exactly $21$ , so the scenario is achievable.

The answer is $\boxed{\textbf{(B)}\ 36}$ .

Review in the AoPS books

Intermediate Counting & Probability, Chapter 13: Events With States (§13.4)
Introduction to Algebra, Chapter 6: Ratios and Percents
Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Algebra, Chapter 7: Proportion (§7.1, §7.2, §7.3)