7
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Place numbers 1 to 64 in the cells of this 8 x 8 board in such a way that consecutive numbers occupy neighboring cells (either vertically or horizontally). Shaded cells must be occupied by prime numbers.

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4
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Here's the solution I got.....

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7
  • $\begingroup$ Interesting. So it seems there are many different solutions $\endgroup$
    – Dr Xorile
    Sep 24 '21 at 19:28
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    $\begingroup$ @DrXorile I wrote a program to check, there appear to be 4 solutions: pastebin.com/UeDRbX5J $\endgroup$ Sep 24 '21 at 20:14
  • $\begingroup$ Me too. Agreed! $\endgroup$
    – Dr Xorile
    Sep 24 '21 at 21:07
  • $\begingroup$ I think there are 4 solutions with a Hamiltonian path and only one solution with a proper tour (where values 1 and 64 are neighbors). $\endgroup$ Sep 29 '21 at 5:47
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    $\begingroup$ @BernardoRecamánSantos See yetanothermathprogrammingconsultant.blogspot.com/2021/09/… for a detailed discussion how I arrived at this conclusion. $\endgroup$ Oct 1 '21 at 22:44
5
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The start has to be:

step1
There's only a handful of primes that are separated by 2. And, apart from 3-5-7, there's no 3 odd numbers in a row that are prime (well-known result, but easily verified up to 64). So that shows us this:
step2
Note the corners have to be connected to the two squares on either side. This means the top-left and bottom-right corners can't be looping back onto the shaded areas (that would be two primes separated by 1). And the bottom left is one of the 5 composites in a row. The lack of 2 primes in a row add a few more restrictions to the path, which I've shown as a red line (i.e. the path can't go through the red lines).
Next I took an intuition that the bottom left must surround the 47-53-59 sequence (with 53 on the isolated shaded spot). I tried a few ways to get the end sequence (which needs 59-61-and the final tail of 3 composites), and eventually hit on using up the space on the right (I almost discarded it because it's not unique). But doing this and the rest fell in place:
final answers

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