9
$\begingroup$

Three friends, Charles, Hugh, and Freddy, played a one-month tennis tournament, just one match between two of them every single day during last December. During the tournament, whoever lost the day's match took the next day off, while the other two friends played against each other. At the end of the month, Freddy had played 17 matches, including those on his birthday (22nd December) and Christmas Day, both of which he won. Hugh, on the other hand, played one fewer game than Freddy, and also won one fewer game than he did.

Who won the match on New Year's Eve, and who lost it?

$\endgroup$
1

2 Answers 2

3
$\begingroup$

On New Year's Eve,

Hugh played Charles

and the winner was

Hugh

Explanation:

There were 31 matches, so in total everybody played 62 times. Freddy played 17 times, and Hugh one less, so 16 matches. That means Charles has to play 29 matches, almost the entire month; he'll miss only two matches. That means he can lose at most three times, including December 31st, and win the rest of the (at least) 28 matches. Freddy wins at least two times, so Hugh at least once. Together (28+1+2), this counts for all of the wins, and as Freddy won on the 22nd and 25th, Hugh must've won on the 31st, defeating Charles on that day.

$\endgroup$
0
2
$\begingroup$

On New Year's Eve

Freddy beat Charles.

Like @Glorfindel said,

There were 31 matches, so in total everybody played 62 times. Freddy played 17 times, and Hugh one less, so 16 matches. That means Charles has to play 29 matches, almost the entire month; he'll miss only two matches.

The only reasons not to play on any given day are:

either you lost the day before or it is the first day and you just happen to not play.

Given that Freddy won on the 22nd and the 25th,

both of those events must correlate to Charles missing a day. Either Freddy beat Charles on that day and played Hugh the next day or Hugh beat Charles the day before and then Freddy beat Hugh.

Therefore

Charles did play on the first day and there were two matches between Hugh and Freddy. Since their win counts are off by one, whoever it was that beat Charles in each of the two upsets was beaten by the third player the following day, keeping their win counts even. The final game on New Years Eve saw Freddy get his third win.

$\endgroup$
2
  • $\begingroup$ Given that Freddy won on the 25th, and therefore played again on the 26th, under what conditions can he also play on the 31st? $\endgroup$ Nov 2, 2021 at 22:19
  • $\begingroup$ He would have to win on the 26th as well and then lose on the 27th, but that already gives him 3 wins and he can't have 4. So Glorfindel's answer was right except for the total win counts (C 26, F 3, H 2) $\endgroup$
    – cap
    Nov 2, 2021 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.