BdMO National Higher Secondary 2019/10

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samiul_samin
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BdMO National Higher Secondary 2019/10

Unread post by samiul_samin » Mon Mar 04, 2019 9:00 am

Given $2020\times 2020$ chessboard, what is the maximum number of warriors you can put on its cells such that no two warriors attack each other.
Warrior is a special chess piece which can move either $3$ steps forward and one step sideward and $2$ step forward and $2$ step sideward in any direction.

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samiul_samin
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Re: BdMO National Higher Secondary 2019/10

Unread post by samiul_samin » Sat Mar 09, 2019 11:40 am

samiul_samin wrote:
Mon Mar 04, 2019 9:00 am
Given $2020\times 2020$ chessboard, what is the maximum number of warriors you can put on its cells such that no two warriors attack each other.
Warrior is a special chess piece which can move either $3$ steps forward and one step sideward and $2$ step forward and $2$ step sideward in any direction.
Actual problem is more interestingly described.
BdMO National Higher Secondary 2019 P10
In chess, a normal knight goes two steps forward and one step to the side, in some orientation. Thanic thought that he should spice the game up a bit, so he introduced a new kind of piece called a warrior. A warrior can either go three steps forward and one step to the side, or two steps forward and two steps to the side in some orientation.
In a $2020\times 2020$ chessboard, prove that the maximum number of warriors so that none of them attack each other is less than or equal to $\dfrac 25$ of the number of cells.

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samiul_samin
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Re: BdMO National Higher Secondary 2019/10

Unread post by samiul_samin » Thu Mar 14, 2019 8:57 am

Given Diagram:
2019-03-14 08.51.48-1-1.png
2019-03-14 08.51.48-1-1.png (30.92 KiB) Viewed 486 times
If we put the warriors at $4k+1$ row,we will get the number of total knights as the question but how can we prove that this the the lowest number of warriors?

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