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Re: DI doubt by Parvesh Angrula - Monday, 3 September 2012, 02:57 PM

let the number of players who received bye in first round be B. let the total number of players be X. so in the first round (X-B) players will play and half of them will loose and will be out of the tournament as no other players received bye or walkovers. so at the end of first round, players remaining in tournament will be (X-B)/2 + B = (X+B)/2 similarly at the end of second round number of players remaining will be (X+B)/2^2 and so on at the end of 9th round, number of players remaining will be (X+B)/2^9 =>this value should be equal to 1, as the winner is only person remaining in the tournament after final (9th) round, so X = 2^9 - B = 512 - B so 533 is not possible. also 199 and 133 is not possible because in these cases number of byes (B) will come out to be more than total number of players (X) so only possibility is X = 487 and B = 25 for the second option we will simply look at the number of players who lost in each round. in first round 487-25= 462 players played, so 231 lost and 231 won in second round 231+25=256 players played and 128 lost (including all 25 byes) and 128 won in the third round 64 won and 64 lost (no byes), so all these lost players also won in both first and second round but lost in third round. So players who won exactly 2 matches is 64. answers 1) 487 2) 64 Show parent | Reply

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