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 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?
- A)
- B)
- C)
- D)
- E)
Answer
B
Key insight
With 3L players the game count is C(3L,2), and wins split 5:7, so the total must be a multiple of 12; only 36 fits both.
Solution
Let there be left-handed and right-handed players, in all. Every pair plays once, so the number of games is
Trying gives games; only appear among the choices.
Each game produces exactly one win. If right-handers won games, left-handers won , so the total is . For this to be an integer must be a multiple of , and then the total is a multiple of . Of , only is divisible by .
Sanity check for : nine players, games, left-handers win and right-handers win . The three left-handers play games among themselves and against right-handers; winning all plus their internal games gives exactly , so the scenario is achievable.
The answer is .
Why this works
Two independent structural facts, "games " and "wins are in ratio ," each restrict the total; intersecting the restrictions with the answer choices leaves one survivor. In tournament problems, always equate total wins with total games, and use divisibility from any stated ratio.
The trap
Forgetting that every game produces exactly one win, so total wins equal total games, or skipping the check that 3L(3L-1)/2 matches a choice.
Common mistakes
- Forgetting that every game produces exactly one win, so total wins equal total games, or skipping the check that 3L(3L-1)/2 matches a choice.
- Interpreting " more" as a ratio of between players rather than between wins, or as left-handers winning of games.
Techniques
Use the answer choices (mod checks, size, form) to eliminate or select · Bound the quantity above/below or estimate to pin it down