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A game is played betwn 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 players as winner, how many players are played in first round?
Read Solution (Total 13)
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- 8th round=2 players
7th round=2*2=4
6 =4*2=8
5 =8*2=16
4 =16*2=32
3 =32*2=64
2 =64*2=128
1 =128*2=256 players
for this type of problems try to solve from last round - 14 years agoHelpfull: Yes(18) No(0)
- 2^8=256
- 14 years agoHelpfull: Yes(7) No(0)
- ans: 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(6) No(0)
- rounds players playing winner
8 2 1
7 4 2
6 8 4
5 16 8
4 32 16
3 64 32
2 128 64
1 256 128
so total number of players played in first round is 256 - 14 years agoHelpfull: Yes(5) No(3)
- at the starting the no. of player will be 2^8=256
- 14 years agoHelpfull: Yes(3) No(0)
- 256->128->64->32->16->8->4->2
No of players is 256 in 1st round - 14 years agoHelpfull: Yes(3) No(0)
- Round No. of Players
8 1
7 2
6 4
5 8
4 16
3 32
2 64
1 128
answer=128 - 14 years agoHelpfull: Yes(3) No(15)
- last round only 2 player
in second last " " 4=2^2 "
in 3rd last " " 8=2^3 "
.............................and so on
here total 8 round
hence 2^8 ans - 14 years agoHelpfull: Yes(3) No(0)
- 256 player
- 14 years agoHelpfull: Yes(3) No(0)
- sorry it's 256
- 14 years agoHelpfull: Yes(3) No(1)
- 8th round would have been played by 2 players to declare 1 as winner;
which implies ,,in 7th round 4 players played to declare 2 as winners;
6th round=>8 players
...
.
.
1st round=>256 players
so the soluton is=>256 - 14 years agoHelpfull: Yes(2) No(0)
- 2^8/2 =128
- 14 years agoHelpfull: Yes(0) No(13)
- 128
- 14 years agoHelpfull: Yes(0) No(13)
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