Brain-Teaser 456: [Chess competition]
From The Sunday Times, 8th February 1970 [link]
The organiser of our lightning chess competition works on a strict timetable; he allows ten minutes a match, and makes no concession, in his planning, for draws or protracted games.
Last year having five chess boards available for use, he arranged an American tournament (in which every competitor played one match against every other player). This year there would be exactly twice as many competitors as last year if one man hadn’t recently dropped out; and only one chessboard is available.
He is therefore organising a knock-out competition: and he reports that his total time allocation will be precisely the same this year as it was last year.
How many competitors are there this year?
This puzzle was originally published with no title.
[teaser456]
S2T2 
Jim Randell 1:56 pm on 27 February 2019 Permalink |
For n players there are T(n − 1) = n(n − 1)/2 matches in a tournament where every player plays every other player.
So, with 5 simultaneous matches this would mean the number of allocated time slots would be:
In a knockout competition with m players there are (m − 1) games. (One player is eliminated in each game, until there is just one player left).
So with only one match going on at a time, the total number of allocated time slots would be:
And we are told that the number in the knockout tournament is m = 2n − 1.
So, equating the times:
So last year there were 20 competitors. This year there are 39.
Solution: This year there are 39 competitors.
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