A game is played between 2 players and one player is declared as winner. All the winners from first round are played in second round. All the winners from second round are played in third round and so on. If 8 rounds are played to declare only one player as winner, how many players are played in first round?
a) 256 b) 512 c) 64 d) 128

• ans: a (256)
  Start from final round. For every round players double in a knockout game. So 8th round(final round)=2 players. Now 7th round double it=4 players. In that way calculate until 1st round. you get 256. or simply 2^n , n= no. of rounds. 2^8=256
• 14 years agoHelpfull: Yes(42) No(0)
• last match(=8th) 2 players
  2ndlast " (=7th) 2^2 "
  3rdlst " (=6th) 2^3 "
  . " (=5th) 2^4 "
  . 2^5
  . 2^6
  . 2^7
  finally(=1st) 2^8
  =256 players
• 14 years agoHelpfull: Yes(19) No(2)
• let x is the total no. of player played in 1st round.so x/2 player can play in 2nd round and x/4 for 3rd round & x/8 for next and by this way x/128 player can play for 8th round and after this round only 1 is winner.bcoz the game is played between 2 players & 1 is declared as winner & all winner is selected for next round.so only x/256 player can say as a winner after 8th round.so x/256 is equal to 1.x=256.so 256 will be the ans.
• 14 years agoHelpfull: Yes(8) No(1)
• 2^8=256....if we divide 256/2....in dis way upto 8 times then we will get at the end only 2 players and finally 2/2=1 and only 1 will b winner..
• 14 years agoHelpfull: Yes(4) No(1)
• 2^8=256
• 14 years agoHelpfull: Yes(2) No(1)