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# Tag Info

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A solution with different sizes of rectangles.

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Proof with brute-force procedure. Here I used fact that we for sure can cover with 16 pentominoes, so I tried to cover half with 8 or less and then see if two such half-covers cover the whole board. It takes about 15 seconds on my PC to get the answer. #include <iostream> #include <vector> const int kHalfUpperBound = 8; const int kSide = 8; ...

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I have found that solution (thanks Python !) : Here is my code if you're curious !

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The patterns are: Column 1 Column 2 The odd ones out being:

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I used a computer, and found the following solution: You can move $16$ or $17$ to the first group, but not both. Those are the only three solutions according to my program, other than permuting the groups. Note that $24$ cannot be added to any group.

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It seems to me that The way the murdering works,

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I can prove that the answer is exactly Several people, including Jaap Scherphuis, have shown that the square can be covered with this many pentominoes, so it only remains to show that at least this many pentominoes are needed. (A matching lower bound). Let us start with the magic board given by A. Rex: As a first lower bound, However, Therefore, we can ...

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Assuming the order of operation is preserved (this is significant only for the first line) and all numbers must be integers So, the only solution with all numbers positive is Explanation:

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Here's another proof of the lower bound in Sriotchilism O'Zaic's answer.

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I think the answer is as folows Reasoning

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This is probably not the minimum, but it is an improvement to the answer from @Sriotchilism O'Zaic: with the sizes In 9 parts: In 8 parts: In 7 parts: In 6 parts: In 5 parts: I used the $18$ pieces from the linked puzzle as my starting point and

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Assuming order of operations is not perserved (multiplication and addition occur from left to right, top to bottom in the column/row), the solution is:

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For completeness, all the solutions, excluding rotations and reflections.

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The problem might be similar to this problem: Given a multiset M of rational numbers where the sum of M equals 1, this is "n-distributable" for an n n∈N, if there exists a partition M1⊔...⊔Mn of M, that the sum of each subset Xi equals 1/n. We would like to find for a fixed k (k∈N) the minimal possible cardinality of a (a1,a2,..,ak) distributable multiset. ...

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People have given some good upper bounds, how about a lower bound. However we can improve this ... However we can improve this ...

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This is a solution with 15 pentominos. Here is a proof that this is optimal. Consider the top and bottom rows, left and right columns, which form the border of the area to cover. Most of the time a W-pentomino covers one or two cells of the border. There are only two ways for a piece to cover more than two border cells. These occur only at the corner of ...

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Here is a solution with 16 pentominos:

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The X-pentomino tiles the plane, so that tiling is a good way to start. There are two ways to cut an 8x8 region out of that tiling. If one of the 4 central squares of the 8x8 region has an X centred on it, you get this or else you get this The latter can be easily improved by replacing the ones at the edges to give this A different way to get the same ...

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I’m thinking: The solution:

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And here are a few 2x14 solutions:

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Simlar to Dr Xorile, I think there are many solutions. Here are several:

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Here's 2 fundamentally different solutions: I suspect there are many because these are literally the first two things I tried as I was playing around and in both cases I was able to just shove the pieces in and get a solution. I did it in the numerical order shown.

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D is equal to: Explanation:

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Here is another solution with the same number of squares as @msh210's: This looks very different to @msh210's, and has the nice property that Furthering on from that:

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The solutions:

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One very efficient way to separate areas by squares is to put them in diagonal lines. This utilises the diagonal, which is the longest dimension of the square. Also, you can divide the plane into separate X-pentominoes (which have the maximum allowed connected area) by using diagonals of squares only. That pentomino pattern has the annoying property that ...

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First, this image shows examples of translation that preserves the summed groups. As mentioned in comment, there are $12$ equivalent arrangements in this class. Here is a solution set: There are seven other solution sets with primes less than 100,000 and countless more with larger primes.

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My best, but is it optimal?

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But I don't know whether that's the least.

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A solution: Proof that this is the maximum:

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This can be done with a method similar to the one I used in the prequel question. In fact, The numbers are displayed below in base 13 to keep the square-ness. (A for 10, B for 11, C for 12 and D for 13) Rotations, reflections, and permuting rows would give other solutions.

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Let's first work out the sizes of the rectangles. It turns out every combination of these shapes is possible.

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This is, Method:

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If I counted correctly and didn't miss any, there are arrangements. There is no particular method to this - just going through all possible combinations of rectangles, and then trying all arrangements. EDIT: An alternative method which does not rely on going through quite so many cases is as follows:

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A simple bound in the more general case:

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Rawrdon Mamsay needs to stay at least days to identify the head chef, and days to identify the cooking order. Reasoning: Rawrdon Mamsay samples the following meals: Rawrdon Mamsay cannot complete his task in fewer days:

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[EDITED to add:] Seeing Magma's answer, posted about a minute after mine, I realise that the question asked two questions and I've only answered one of them. I shan't modify my question to answer the first as well because Magma's answer already does that perfectly well. I am not certain that I've understood the explanation in the question quite correctly, ...

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I think the following would work Strategy

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Again got it on the first try...

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It would appear that I further remark that

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Is this what is meant? Looks trivial though...

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Short and Concise solution, Finally,

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I was having a slow work day, so I fired up Blender and made this: In 13 hops, the block of 9 pegs can be moved two places down and to the right. By repeating the process two more times, the pegs can be moved to the bottom right corner.

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EDIT EDIT 2 Made a computer program to look for a solution with fewer moves. It managed to improve my previous solution so it now only takes Here they are: I had to make some assumptions to get a solution within a reasonable time, so I'm not absolutely sure this is the minimum. I'd love to see the minimum if this is not it!

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I would say that the Because,

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So, there is actually an approach by hand (without calculating via a computer program or creating a $10 \times 10$ table of a recurrence relation). Observe that to get a distance of $10$ by using planks of length $2$ and $3$: Thus, from $(0,0)$ to $(10,10)$, the number of ways — including going into the whirlpool — is: Now, we may then count ...

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Using the following program I get let result = 0; go(0, 0); function go(x, y) { if(x === 10 && y === 10) { result++; return; } if(x > 10 || y > 10 || (x === 5 && y === 5)) { return; } go(x + 2, y); go(x + 3, y); go(x, y + 2); go(x, y + 3); } console.log(result);

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