5. For a table boxing tournament in Yoshkar-Ola, 52 fighters have gathered. It is known that all have different levels of strength and in the game, the stronger one always defeats the weaker one, with one exception: the weakest fighter is an inconvenient opponent for the strongest and always defeats him. The real strengths of the fighters are unknown to the organizers. Can the organizers identify the strongest fighter in no more than 64 matches?
Solution. Answer: they can.

Let the strongest fighter be $S$, and the weakest $W$. Divide the fighters into 13 quartets. In each quartet, conduct the following matches: by pairing the quartet, determine the winners in the pairs (let $a>b$ and $c>d$), then determine the winner in the pair of winners (let $a>c$). Fighter $a$ has defeated two others, so $a \neq W$. Therefore, $b \neq S$ and $c \neq S$. Compare $a$ and $d$. If $a>d$, then $d \neq S$, so $a$ is the strongest in the quartet. Then move on to the next quartet. If this situation repeats in all quartets, we will have eliminated 39 fighters in $13 \cdot 4=52$ matches, and among the 13 remaining fighters, there will be no $W$. By pairing the remaining fighters in any way and eliminating the loser, we will find $S$ among them in another 12 matches.

Suppose, however, that in some quartet, the fourth match resulted in $ac>d>a$. Clearly, this is only possible if there is a cycle involving $S$ and $W$: specifically, $c=W, d=S$ or $d=W, a=S$. Compare $b$ and $d$ ($b \neq S$). If $b>d$, then $a=S$, and if $b<d$, then $d=S$. Thus, we have found $S$ in no more than 53 matches.