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Here is a solution with a constant product of which I think is the minimum possible: Some partial progress for a lower bound on the product: This leaves only a few possibilities for improvement:

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I found a solution trying to minimize prime factors. And finding a balance between the minimum value and minimum number of factors. I noticed that some fields are connected, in that multiplying one of them results in the multiplication of fixed other fields. There are three such patterns: -C, D, H, L, and M (and any mirror image of that) -A, E, G, L, M (...

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Edit: In an effort to find the minumum, here is a much smaller solution in which the mutual product is Solution As MKBakker pointed out we could further reduce this by dividing each of the entries 4,8,16,96 and 192 by 2 to get a mutual product of although they have subsequently improved on this.

14

There are a number of ways to do this. An easy strategy:

9

To do this, Finally, I can give you the Python code if you want. Not sure it can be spoilered (I've never learned how to have multi-line spoilers). Python Code

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3

However Why? so whatever Bob can arrange these cards in the best case scenario, Let's make it more complex; so even it is 0 to 100, Ann will

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To formalize user61579's answer, a simple reinterpretation of the rules is that, defining a pass as a move from the $1$ all the way around the sequence, every pass, every second $-1$ becomes $0$ Then, it is trivial to show that the state of a length-n sequence after each pass corresponds to the first n columns of this infinite table: 1 -1 -1 -1 -1 -1 -1 -1 ...

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Code to find the pattern: https://ideone.com/O5S4Qu (The third number printed out on each line is the winning move if the player to move is in a winning position)

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We have a winning strategy for: The first move is:

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For the first question, there is an easy pattern that we can detect. First, all of the nice jumps will come from the n-1 position for an array of length n. As for how many, I'll print the first few and then try to point out the pattern 2: 0 3: 0 4: 1 5: 0 6: 2 7: 1 8: 1 9: 0 10: 3 11: 2 12: 2 13: 1 14: 2 15: 1 16: 1 17: 0 2 and 3 are basically special ...

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Here's a little Python program to test it yourself: https://repl.it/repls/ScientificIdenticalPixels And here's C++ code written by user @im_so_meta_even_this_acronym https://ideone.com/SfMHqC

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