2023 AMC 12A Problem 13

Question

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?

15

36

45

48

66

Answer: B

Concepts:combinations divisibility

Difficulty rating: 1660

Small hint:

Let there be

L

left-handed and

2L

right-handed players, so

3L

players total

Big hint:

If right-handers win

R

games, the total is

2.4R,

so the total must be a multiple of

12

Solution:

Let there be

L

left-handed and

2L

right-handed players, for

3L

players and

\binom{3L}{2}

games total.

If right-handers win

R

games, left-handers win

1.4R,

so the total is

2.4R=\tfrac{12}{5}R.

For this to be an integer count, the total number of games must be a multiple of

12.

The left-handers can win at most every game involving at least one left-hander, namely

\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}.

On the other hand, they must win

\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8

games. Comparing these quantities gives

21L-7\le 20L-4,

so

L\le 3.

Testing

L=1,2,3

gives totals

3,15,36;

only

36

is a multiple of

12.

It is achievable: the

3

left-handers can win all

18

mixed games and their

3

internal games, giving

21=1.4\cdot 15

wins.

Thus, the correct answer is B.