Peter, Paul, David join a table tennis tournament. On the first day, two of them were

randomly chosen to play a game against each other. On each subsequent day, the loser of the game on

the previous day would be benched and the other two would play a game. After a certain number of

days, it was found that Peter had won 22 games, Paul had won 20 and David had won 32. How many

games had Peter played?

## AIME

- asif e elahi
**Posts:**185**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: AIME

Every time Peter loses, he doesn't get to play in the next game. So we can (almost) make an one-to-one map between Peter's losing and not playing days. i,e if $x=$ number of games Peter lost, then $x = (22 + 20 + 32) - 22 - x \pm 1 \Rightarrow x=26$. So Peter played $26 + 22 = 48$ games.