Question

question

Answer 1

There are 64 players in a knockout chess tournament ,
how many total matches would be played ,
if all matches are decisive ?

Answer 2

keep dividing by two

Answer 3

Nin, Iʻll divide you by two

Answer 4

32,16,8,4,2,1

Answer 5

or log2(64)

Answer 6

Yeah lets not jump to logarithims

Answer 7

\(\Large \log_{2}{64}=6\)

Answer 8

c;

Answer 9

but

Answer 10

how many total matches would be played

Answer 11

answer is not \(\Large 6\)

Answer 12

in the first event, there are 32 matches since 32*2 = 64
in the second event, there are 16 matches since 16*2 = 32

etc etc

Answer 13

Initially there are 64 players. There will be 32 matches between 2 people, and only 32 will move to the next round(say) and rest 32 get eliminated. Then 32 will further get reduced to 16..and so on..so the total answer is 32+16+...

Answer 14

\[\sum_{i=1}^{5} 2^i\]

Answer 15

simple methodology works :)
thank you for your time

Answer 16

\(\Large \sum_{i=1}^{5} 2^i=62\)

62 is not the answer

Answer 17

62 is not the answer because there are 63 goddamn matches if you just followed what I said and added all the matches instead of doing fancy summation and logarithms

Answer 18

+1 for final match :)

Answer 19

\(\Large \sum_{i=0}^{5} 2^i=63\)

Answer 20

okay it can also go like this->
we have 64 players
we just make 2 players play and then the winner goes ahead and competes with next player and the winner here goes ahead like this.. so

if we have \(n\) players then total matches be \(n-1\)
:) so no need to divide and add
its just \(n-1\) so \(64-1=63\)

Answer 21

here's a lesson
http://regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm