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  • $\begingroup$ This is a very nice generalisation. It means we can find solutions for a grid of any size. $\endgroup$
    – Dmitry Kamenetsky
    Commented Aug 3, 2022 at 23:40
  • $\begingroup$ @DmitryKamenetsky I don't think that's true. This answer requires a set of moves which produce 1 on all cells. Such a move does not exist for any grid. E.g. a 2x3 grid cannot be filled with all 1s. $\endgroup$
    – Falco
    Commented Aug 5, 2022 at 8:47
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    $\begingroup$ I especially like this answer, because it found the solution (16) without actually searching for any specific primes. $\endgroup$
    – Falco
    Commented Aug 5, 2022 at 8:48
  • 1
    $\begingroup$ @Falco A $2\times 3$ grid can be filled with all-one by making moves in two opposite corners. With small modification, this method also works if one can produce an "all-equal" result, which is equivalent to saying that the "all-one" vector lies in the $\Bbb Q$-vector space generated by the basis vectors. I have checked every rectangular grid up to $20\times 20$ that this is always true. Perhaps there is a general argument to prove that. $\endgroup$
    – WhatsUp
    Commented Aug 5, 2022 at 14:56
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    $\begingroup$ @WhatsUp, I think that argument requires negative moves? But that's fine, just crank up the binary representation grid with more moves. $\endgroup$
    – Ben Reiniger
    Commented Aug 5, 2022 at 16:00