# Tag Info

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I wrote a computer program and it showed that $18$ moves is the optimum. Here is one such solution: Oddly enough, even if you relax the condition of alternating white and black moves, it cannot be done in fewer moves. For $3\times3$ the optimal number of moves is $16$. Without the need to alternate moves the optimum is $14$ moves, for example just by ...

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You can do better than the greedy algorithm. With coins of value you can get N = I wrote a search program for this but it isn’t going to finish searching the entire space in reasonable time, so I don’t know whether that’s optimal. Here are all the optimal solutions for up through seven coins:

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I can do a little better, using coins of I can pay amounts up to (and including) (without change). Explanation: One can ask what $N$ is as function of the # of different coin values and maximum # of coins in a single payment.

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The answer is: Here's the way you get that number:

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The answer is 7, It's a simple extension of the Magic Calculator trick. Each card represents a bit, with the upper left index value being the value of that bit. To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have ...

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The one-dimensional problem can be solved as follows: Using chessboard coordinates:

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I can improve a little bit further using the coins With those coins I can express (with at most 8 coins, and no change) every non negative integer up to (and including) I'm not sure this is the definitive answer, as I was able to outperform the greedy algorithm for a smaller number of coins. Explanation of how the program works:

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@hexomino has found optimal solutions with a restricted search space. Theorem: Proof: Therefore Theorem Proof Therefore

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I believe I've found a solution worth 16 points.

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Sum Product Reasoning

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I believe I can open it in The reasoning behind it is:

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I can ensure that I receive at least To get this amount of coins, I could apply the following strategy: My friend can also ensure that I do not get more coins than this in the following way: So it appears that I should have chosen a fairer game to distribute the coins...

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This is like a reverse puzzle to the well-known Here in reverse: A remark as a thinking outside the box solution:

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It takes Suppose Now

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Between 10 friends, there are only 45 unique pairs. Each friend will participate exactly 9 times, which means after 45 rounds everyone will be level 10. Since it's impossible to get an item on the first & last run, there is a theoretical maximum of 43 items you can get. However, due to some restrictions, it seems to me like it's impossible to get this ...

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The optimal solution is This set of coins allows N = This set of coins is given in this paper: Some Extremal Postage Stamp Bases, by Michael F. Challis and John P. Robinson. This paper was found by @alephalpha.

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Here's the solution, and a proof of its optimality: First, an important fact: Now, analysing the grid: Constructing the solution:

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Got 19 by moving around... might be possible to do better:

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I think the answer for $14 \times 14$ is Achieved as follows While the best I've achieved for $9 \times 9$ is Achieved as follows

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I may need some clarification on counting but I think I've found a solution in which the landlord needs to repair 38 squares while filling just 3 other apartments with water. Solution Partial Reasoning

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The smallest slice would be: The procedure I followed was: This leads to: And the students get:

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Fact 1: Fact 2: Proof: Exhaustion... Maximizing possibilities: Notation: (x,y,z) : highest possible value Possibility 1: (1,2,?) Possibility 2: (1,3,?) Possibility 3: (1,4,?) Overall:

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A solution in: Borrowing from @Jaap:

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My solution (but see my new, second answer): EDIT: thanks to @mlk.

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@Oray congrats this question is glorious!

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