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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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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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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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[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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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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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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The solutions:
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My best, but is it optimal?
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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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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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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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A simple bound in the more general case:
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