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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, ...

1

I believe this qualifies:

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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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I can do it in Note: Continuing... Assume the worst case. Any other results would obviously be easier to solve. For any of these options: Thus, we have completed the process and identified card 8 in at most Note, that as a side-effect of identifying card 8, we also identified cards 12, 11, 10, & 9. There might be a more efficient option to ...

5

I can do it in:

2

Nobody has a offered a bishop construction yet. Here is one, though imperfect: White to move and mate in 3 Solution: It is not great because The difficulty to build a problem with BB or RR promotions is that

1

Short and Concise solution, Finally,

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Promotion to rook, 5 pieces (note this may not qualify)

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Currently on mobile, so I'll have to provide the full list of moves later, but I think I have it ending with: The full list of moves:

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Promotions to knights, 8 pieces White to play and mate in 3. Solution:

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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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Assuming "transportation cost" means sum of distances to each of the three roads, and the side of the equilateral triangle has length $1$:

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

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

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For promotion to queen, a solution with four pieces: Best play is For promotion to knight, one with nine pieces: The solution here is

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Asnwer Diagram explaination... Therefore my answer is... The general formula is that with n straight lines we can form as many as if n is ODD. if n is even If there are m straight lines at right angles to n straight lines, m being less than n, the

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I got: Counting: As a picture

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I assume we have to draw exactly 111 squares and the lines are finite. My answer: How I find that: class Program { static void Main(string[] args) { Console.WriteLine("***********"); printGrids(111); Console.WriteLine("***********"); printGrids(112); Console.WriteLine("*******...

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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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I think the best you can do is As follows Counting

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