# Tag Info

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I found

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Part A: Part B: Part C:

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I found an answer im satisfied with : Reasoning and variations : This puzzle was fun to solve, I really enjoyed it. I still don't know if this is the best solution. With less pieces, checkmating gets really hard. PS : My solution is computer-free, and im not good enough at programming to be able to verify my solution. If I missed something or if there is ...

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I believe the answer is The reason being Maybe this is wrong, but I'd love to hear your thoughts.

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Maybe I'm missing something, but it seems that But I'm not certain.

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I got This was my try. I haven't counted the steps yet but I will do it later.

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I tried another path : i got Here's my path (sorry for the very bad paint editing. There's a yellow mark every 10 moves) I tried to make the two 'back and forth' sections as clear as possible, but the left one is still hard to see. From the bottom, I head to the boost, then I grab the X and i go back to the boost. This was done without a program so there ...

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The best I could do is The path I took was Red is the main path, and blue indicates a short trip to grab an X and get back onto the path. I do not know for sure whether the solution is optimal.

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As an example, say we have numbers 1, 2, 3. Then, we find the sum of the three numbers, which is six. Then, we take the numbers and lay them out in a row. Then, we ask the sum of two slips. Then, we ask for the sum of another two slips. The one with the greater sum has the biggest number in the group. The one with the smaller sum would have the smaller ...

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The answers for $M$ are somewhat surprising! I expected smaller numbers. $k=0$ Minimum Maximum $k=1$ Minimum Maximum $k=2$ Minimum Maximum

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You can get at most This is possible to achieve: So now we've reduced the problem to the variant where Here's an optimal strategy: This is optimal because:

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Without restriction of only checking consecutive papers The general solution for $n$ would be: Visualization: With condition of only checking consecutive papers I believe this is the general equation for $n$: For $n > 2$:

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I have improved the queens solution or some more

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The most moves possible for both sides in a position where all 32 units are still on the board is 164. This answer was stolen from @GloriaVictis's answer to this CSE question. Nenad Petrović, The Fairy Chess Review 1946

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Via integer linear programming, the maximum for knights is The maximum for queens is at least Other maxima are

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Since this is a puzzle, one could interpret the condition "different from the cube which was already in the box" to refer to just the first cube (subsequent cubes weren't already in the box when the "After doing this" part starts). In that case, you could have as many cubes as the box can hold, so long as they were all different from the first cube. If you ...

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Here are my own solutions I put together. With the exception of the Rooks solution, they have already been equaled or surpassed by Daniel's answer, so I'm just posting them for comparison. Bishops: Rooks: Knights: Queens: Kings:

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First, we find the number of ways possible to paint one side of the cube: 1, because the rest will repeat For two sides: 2 For three sides, there is 2 For four sides, there is 2 For five, there is only 1. For all sides, I'm not sure if this is needed for the final solution but it has only 1 way. Thus, when you add $1+2+2+2+1+1$, ...

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Bishops Knights Rooks Queens Kings

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Initial thoughts Update 1

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If I am correctly understanding the problem statement (see my comment above) then I think this does it:

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How many rabbit maximum would Roger have after a year from the adoption of his pets? How exactly can I find the number of rabbits in this situation? For this situation, I would just write it down. There are only four steps to take (4 quarters in a year), thus not much work to write it down. The growing rate is exatcly the same for the female as for the ...

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As the number of rabbits added is thrice that of the original, the population of rabbits is quadrupled every 3 months. As such the growth can be modelled by $4^{(n/3) + 1}$, where n is the time in months after the first adoption. This can also be done iteratively as such: 0 months: 2F, 2M 3 months: 2F, 2M + 2(3F + 3M)...

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The last episode was difficult, with a few twists and turns, but this selfmate left little doubt: the answer is 18 moves. Explanation It's clear that we must manipulate Black's pawn structure somehow, because there's no way a mate can come out of the current structure. The fact that such a manipulation must be forced, i.e. Black's only legal move, ...

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Put on one side of the scale the 250 gr weight then put on the other side the 500 gr weight. Put the 5000gr bag on one side and an identical empty bag on the other side. Start taking sugar from the full bag and put it on the other till the scale is balanced. Then you have 5750gr divided by 2 equals to 2875gr if you take the weights off the scale then you ...

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First, ..and, well, you are done. As for the strategy for finding the solution, look for integer multiples of the weights (and their sums and differences) in the amounts of sugar (and their sums and differences). If you find none, try halving some of the numbers, which is easy to do on balance scales. Here you get two "direct hits" right off the bat, so ...

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In: How?

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Observations A few basic facts I notice: The breakthrough threat of ...b4, cxb4, and ...c3 prevents the White king from leaving the square a1-d4, namely we must rule out a winning plan of running to the kingside and gobbling up the black pawns on g4 etc. The a6-b7 structure keeps Black's king confined, both spatially and temporally, to b8-a7 (or c7). ...

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Wow, that sure is tight. And yes, even though it doesn't look like it at the first glance, the I and T pentominoes are legally connected to each other :-) This took surprisingly long to find and (especially) double check, so I have nothing for the general case with n players.

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Here is a solution with pieces. The colouring of the empty squares ($2\times2$ in white, $4\times4$ in yellow, $6\times6$ in orange, $8\times8$ in red) shows how the pattern generalises to any $2k\times2k$ board. For $8\times 8$ this solution is optimal.

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Non-overlapping:

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Non-overlapping It can be done using As follows: Proof that this is minimal: Overlapping It can be done using Proof that this is minimal:

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A quick attempt using polyominoes: Here is an attempt for the bonus question using polyomnioes

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Looking at one of the tags*, it is not possible to:

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Getting a bit crowded here with empties:

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Here is my attempt and probably optimum if I didnt miss something: It is the idea is simply; In other words,

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