# Here is a knockout formula

Last Updated on 11 October 2011 by gerry

#### a = 2 (n – R)

I always like a good real-world application of mathematics, in this case algebra.

#### The question

If you’re organising a knockout tournament for something and the number of players or teams is not a neat power of two, how many players have to be drawn in round 1, so that round 2 is a neat power of two? I figured this out by solving this pair of equations:

`a + b = n`
`0.5a + b = R`

where `a` is the number of players to be drawn in round 1, `b` is the remainder of players to be drawn in round 2, `n` is the total no. of players and `R` is the largest power of two less than or equal to `n`.

#### Solving the equations

`0.5a + (n-a) = R`
`a + 2n - 2a = 2R`
`2n - a = 2R`
`a = 2n - 2R`
`a = 2(n - R)`

#### An example

91 players enter our tournament, how many must be drawn in round 1?
`n = 91, R = 64`
`a = 2 x (91 - 64)`
`a = 54`
So we draw 54 players in round 1, which produces 27 winners to meet the remaining 37 players in round 2.

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