My answer is 2020*505. That many warriors can be put but I am not sure if we could do better.

meet18 wrote: My answer is 2020*505. That many warriors can be put but I am not sure if we could do better. how?

thegreatp.d wrote: meet18 wrote: My answer is 2020*505. That many warriors can be put but I am not sure if we could do better. how? Put the warriors in the columns mod $4k+1$.

Math1Zzang wrote: thegreatp.d wrote: meet18 wrote: My answer is 2020*505. That many warriors can be put but I am not sure if we could do better. how? Put the warriors in the columns mod $4k+1$. Exactly, that's my idea I think that's the best we could do.

You have to prove it.Idea is not enough.Your idea is correct.

It's not hard as well. In our construction, each cell not containing a warrior is attacked by exactly four warrior. That's the maximum amount achievable with warriors not attacking each other. Thus, we are using the cells most judiciously

This is also Bangladesh National Mathematical Olympiad 2019 Higher Secondary P10.

I don't even understand the question

Holmes777 wrote: I don't even understand the question Just think about the piece knight first and then replace it with this special type piece 'Warrior'. If you put a warrior at the middle of a $7\times 7$ chess board it can attack $12$ squares. Draw a picture of the movements of warrior.

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I think getting 15 out of 15 is tough but approching this one is not so hard.I have tried it for about 1 hour in the examination hall .

This is 1/50 difficulty of IMO because I solved it.

meet18 wrote: This is 1/50 difficulty of IMO because I solved it. Will you please post the full solution?I want to know the solution.

strategy: put the warriors in rows of the form 4K+1 proof:It's not hard as well. In our construction, each cell not containing a warrior is attacked by exactly four warrior. That's the maximum amount achievable with warriors not attacking each other. Thus, we are using the cells most judiciously.(I leave why 4 squares are maximum to the reader)

meet18 wrote: This is 1/50 difficulty of IMO because I solved it. ........................

IMO 2017 P4 is way more easy than this one. See here.

meet18 wrote: strategy: put the warriors in rows of the form 4K+1 proof:It's not hard as well. In our construction, each cell not containing a warrior is attacked by exactly four warrior. That's the maximum amount achievable with warriors not attacking each other. Thus, we are using the cells most judiciously.(I leave why 4 squares are maximum to the reader) I don't understand your wordings. What do you mean by "we are using the cells most judiciously"? Also, I don't think "each cell not containing a warrior is attacked by exactly four warrior" when we are considering the edges. I may misunderstand the question. Please tell me if I really did.

meet18 wrote: It's not hard as well. In our construction, each cell not containing a warrior is attacked by exactly four warrior. That's the maximum amount achievable with warriors not attacking each other. Thus, we are using the cells most judiciously This is a contest problem.You can not just say that it will happen.You have to prove it.

guys I want to share something. Look in many competitions(also in BD) there is a problem related to chess.Can we have a marathon for this kind of problems? In this way we will be familiar with such problems and will be able to solve them. Please share your opinion.

I also want it.I have some nice problems related with chess.I have also made some problems related to chess.

so we do it here or in bdmo forum?

BdMO forum is not so active now(as BdMO National is ended) .So I prefer to do this marathon here.

then we can start

Holmes777 wrote: then we can start In which sub forum?

whichever you like

@Holmes and @Olympus if you want to converse, please add each other as friends and talk in PMs.

Holmes777 wrote: guys I want to share something. Look in many competitions(also in BD) there is a problem related to chess.Can we have a marathon for this kind of problems? In this way we will be familiar with such problems and will be able to solve them. Please share your opinion. yeah I want a marathon-Let's start it!

Marathon is started here.But why full solution of this problem is not posted yet?Is this too hard to prove?

Olympus_mountaineer wrote: Holmes777 wrote: then we can start In which sub forum? I am just trying to post the solution. Sorry if their will be any mistake. Strategy: Put the warriors in $(4k+1)$ columns. Proof 1)We can easily notice that warriors can't attack warriors in his own column. So we can arrange the warriors in above mentioned arrangement. 2) The warriors of $(4k+1)$column can attack warriors of $4k,4k-1,4k-2$ columns . So we can't put the warriors in those cells. So the maximum number of warriors that can be put is $2020 \times 505$

ayan_mathematics_king wrote: Olympus_mountaineer wrote: Holmes777 wrote: then we can start In which sub forum? I am just trying to post the solution. Sorry if their will be any mistake. Strategy: Put the warriors in $(4k+1)$ columns. Proof 1)We can easily notice that warriors can't attack warriors in his own column. So we can arrange the warriors in above mentioned arrangement. 2) The warriors of $(4k+1)$column can attack warriors of $4k,4k-1,4k-2$ columns . So we can't put the warriors in those cells. So the maximum number of warriors that can be put is $2020 \times 505$ The construction is maximal, but is it maximum?

Is there any official/true solution?

I think the maximum number is 1632160.