Training for game throughout 11 weeks, practicing at least 1 game per day and max 12 games per week Proof that there is number of days sequence that equal to 21 games
Answer:
$x_{i}$ sum of number of games until day $i$
$1\leq x_{1}< x_{2}..\leq x_{77}= 12*11 =132$
we need to find $x_{i} - x_{j} = 21$ or $x_{j}=x_{i}+21$ then we can claim that sequence of days sum are 21 games
then we add 21
$22\leq x_{1}+21< x_{2}+21..\leq x_{77}+21=153$
{$x_{1},x_{2},..,x_{77}$}, {$x_{1}+21,x_{2}+21,..,x_{77}+21$} we have 154 numbers. the sum can range from 1 to 153 so there is 154 cells and 153 pigeon
so, $\left \lceil \frac{154}{153} \right \rceil = 2$ we get two equal number in same cell
My question:
- Why 21 was added to the inequality ?
If there is an $i,j$ such that $x_i-x_j = 21$ then $\{x_i,x_j\}\cap\{x_i+21,x_j+21\}$ will be non-empty.
Or, the number of members of $\{x_i,x_j\}\cup\{x_i+21,x_j+21\}$ will be less than the number of sum of the number of members of each set.
Contrariwise, if the number of members of the set $\{x_1,\cdots, x_{77}\}\cup \{x_1+21,\cdots, x_{77}+21\}$ is less than $77+77$ then there is at least one member of each subset that are the same.