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  • $\begingroup$ I'm attempting to solve this with the original solution (dividing into overlapping groups) and have so far shown that four groups is insufficient. Unfortunately, the 5-cube is a little too complex for me, so I might just wait until someone comes along with a more elegant solution. $\endgroup$
    – Auride
    Commented May 30 at 0:06
  • $\begingroup$ @Auride Once you have solved or seen the solution to the original problem, it is just too tempting to try to adapt it. I think that is why people (including myself) find/found it so difficult to find a counterexample to the conjecture. 20 years ago all the maths PhD students at UCL spent a week trying to think of a counterexample and could not. I have seen other groups of similarly smart people come up with the same conjecture in online discussions, and fail to find a counterexample too. I only found this one by reading a research paper on the topic. Once I understood the theory ... $\endgroup$
    – tkf
    Commented May 30 at 0:38
  • $\begingroup$ .... I realised that if you phrase it this way, there is an elementary solution that requires no knowledge of the original problem, or any of the related theory (in fact not knowing is probably an advantage). $\endgroup$
    – tkf
    Commented May 30 at 0:41
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    $\begingroup$ I have a pretty clean solution if there are 5 squares outside the 3x3 grid, or if there are 4 coins outside the 3x3 grid and flipping a coin is optional. Still working on a solution to the original problem. $\endgroup$
    – isaacg
    Commented May 30 at 19:17
  • $\begingroup$ @isaacg Brilliant - I would love to see your solution. Given the context, I think anything less than 7 extra coins deserves to be posted as an answer and upvoted (but not technically accepted). $\endgroup$
    – tkf
    Commented May 30 at 19:37