5. Peter, Paul and David joined 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 take a rest and the other two persons would play a game against each other. After a certain number of days, it was found that Peter had won 22 games, Paul had won 20 games and David had won 32 games. What was the total number of games that Peter had played?
(1 mark)
小明、小強和小輝參加了一個兵兵球比賽。在第一天，在他們之間隨意抽出兩人比賽一場；在之後的每一天，均是之前一天比賽的負方休戰，而另外兩人則比賽一場。若干天後, 發現小明共勝出 22 次、小強共勝出 20 次、小輝共勝出 32 次。那麼, 小明共比賽了多少場？
5. 48
5. Since $22+20+32=74$ and there is exactly one winner each day, we know that the 'certain number' of days in the question is 74 . Hence there are $74-22=52$ days on which Peter did not win - he either lost the game $(\mathrm{L})$, or he did not play $(\mathrm{N})$. These must be equal in number (i.e. 26 each), since if we denote each of the 52 days in which Peter did not win by $\mathrm{L}$ or $\mathrm{N}$ and list them in order, each $\mathrm{L}$ must be followed by an $\mathrm{N}$ (except possibly the rightmost $\mathrm{L}$ ) and each $\mathrm{N}$ must be preceded by an $\mathrm{L}$ (except possibly the first $\mathrm{N}$ ). Hence the answer is $22+26=48$.