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Football and Combi

Posted: Thu Apr 12, 2018 2:32 pm
by Abdullah Al Tanzim
$23$ people, each with integral weight,decide to play football, separating into two teams of $11$ people and a referee.To keep things fair, the teams chosen must have the equal total weight.It turns out that no matter who is chosen to be the referee, this can always be done.Prove that the $23$ people must all have the same weight.

Re: Football and Combi

Posted: Fri Apr 13, 2018 9:54 pm
by samiul_samin
It is not possible if any of the referees is not same weighted.So,every referee has the same weight. There are 23 possible referees.So,it is proved that everyone has the same weight.(Short answer).

Re: Football and Combi

Posted: Sat Apr 14, 2018 12:43 pm
by Abdullah Al Tanzim
samiul_samin wrote:
Fri Apr 13, 2018 9:54 pm
It is not possible if any of the referees is not same weighted.
why?

Re: Football and Combi

Posted: Sat May 19, 2018 8:33 pm
by Golam Musabbir Joy
Let the weights of the peoples are $a_1,a_2, \dots a_{23} $ and $\sum a_i = S$. It is clear that $S - a_i$ is always even. So that $S $ and all $a_i $ nust have the same parity. It is also clear that if $a_i $ is a solution then $a_i + 1$ is also a solution. Let $b_1, b_2, \dots , b_{23} $ be one of the example where $S $ is least and all the $b_i $ is not equal.
If all the $b_i $ is even, then choose another sequence $c_i = \frac {b_i}{2}$ and if all the $b_i $ is odd , then choose another sequence $c_i = \frac {b_i - 1}{2} $ for $1 \leq i \leq $ which is also a solution and $S $ become lessen this time. So all the $b_i $ must be $0$ which implies all the $a_i $ is equal.