# Tag Info

36

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 ...

24

You're

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(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand. First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31. Second step: let's try to build the shortest number possible from these numbers. ...

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The answer is I wrote a Java program to find it:

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Here's an optimal procedure that lets you find out the status of every person after exactly questions: Reasoning: Proof of optimality:

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

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

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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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Glorfindel's answer is sufficient for the main question. To answer the bonus question: Here is an example: To construct this example, As for a starting position,

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So I can't yet prove this is the smallest, but it's at least an upper bound: Reasoning:

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The answer is: Here's the intuition: Finally, here's

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Shuffling the sequence of 5 and the non- overlapping 19, 23, and 29 by trial and error produces:

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I managed to find the 8 in So we are trying to find the 5th largest card in a bunch of 12, by measuring them in batches of four. Here's my strategy: Now we have identified, for certain, a couple of cards we can exclude: A1 and A2 (both have at least 8 cards smaller than them), and B4, C2, C3 and C4 (all have at least 5 cards bigger than them). We also have ...

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You only need Assume the genuine coins weigh $x$ grammes. Credit to Hexomino and Jaap for the corrections!

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

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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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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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The answer is Moreover, First note that Step 1 Step 2 Step 3 Step 4

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Here is a way to do it with four digits Here is a way to do it with three digits

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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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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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EDIT: Some have suggested that the question might be to find exactly 111 squares. In this case, ORIGINAL ANSWER Let's count

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Improvement on the 2-explorer case: First attempt at 3-explorer case, using the same strategy and formatting:

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Improved answer This is similar to another answer but with a smaller area. I worked it out completely independently, then noticed its similarity. The 3" dish is in a different place, and it is not an adjustment based on that answer. It was generated by a C program I wrote for this purpose. It gave my previous answers and has been spitting out smaller ...

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

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Thinking about this problem in reverse is easier, i.e. think about the trees that need to remain. Find the fewest number of trees that can be arranged in five lines with 4 trees in each line. To minimise that number, we need as many trees as possible to do double duty by being part of multiple lines.

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

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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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