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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 ?
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  • $\begingroup$ Becuase you want to prove there is an $x_j = x_i + 21$. And we do that bylooking at the ranges of all possible $x_j$ and at the range of all possible $x_i + 21$ we seeing if there must be some overlap. If the range of $x_j$ is $1$ through $132$ then the range if $x_i +21$ is $22$ through $153$. $\endgroup$ – fleablood Aug 25 '20 at 18:30
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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.

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  • $\begingroup$ why $x_{i}−x_{j}=21$ can't be found in the first inequality ? $\endgroup$ – Mostfa shma Aug 25 '20 at 18:10
  • $\begingroup$ How do you intend to compare all possible $i,j$ pairs to show that at least one is equal? This is a shot-gun approach, that allows us to check the set en-masse. $\endgroup$ – Doug M Aug 25 '20 at 18:11
  • $\begingroup$ assuming $x_{6}=42 x_{3}=21, 6=i, 3=j$ then what are the sequence elements in $x_{i}-x_{j}$? $\endgroup$ – Mostfa shma Aug 25 '20 at 18:17
  • $\begingroup$ They aren't ordered in sequence. You have $x_6-x_5, x_6- x_4, x_6-x_3, x_5-x_4, x_5-x_3, x_4 -x_3$. But we don't just have $x_6$ through $x_3$ we have $x_77$ through $x_1$ so that is $\frac {77*76}2$ possible $x_j - x_i$ and there is no need that we can tell than any of them have to be $21$. The smallest they can be is $1$ and the largest is $131$ so they must double up we have no reason that any of them must be $21$. $\endgroup$ – fleablood Aug 25 '20 at 18:38

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