A standard set of double-six dominoes has 28 tiles with 2 numbers on each side from 0 to 6. Tiles can be placed next to each other if the numbers at each end match. Can you place all the 28 tiles such that they form a loop? Note that the end number must also match the front number to complete the loop.


Note that

the numbers, $0-6$, are elements of the finite field $\mathbb{F}_7$.

So there are

circles like this: $(0)(d)|(d)(2d)|\cdots|(6d)(0)$ for every $d\in\mathbb{F}_7^\times$.

Now just

connect all the circles, and insert the remaining $7$ tiles of the form $(d)(d)$ anywhere you like.

The final answer:

00 01 11 12 22 23 33 34 44 45 55 56 66 60 02 24 46 61 13 35 50 03 36 62 25 51 14 40

  • $\begingroup$ That is correct! Great answer. $\endgroup$ – Dmitry Kamenetsky Oct 24 '19 at 1:13
  • 1
    $\begingroup$ You don't need the finite field, as any set of cycles will do. It is essentially an Eulerian circuit through a complete graph (with a loop on each vertex), and the method you describe is a general algorithm for any graph that admits a Eulerian circuit. $\endgroup$ – Jaap Scherphuis Oct 24 '19 at 4:14
  • $\begingroup$ @JaapScherphuis Of course I understand the interpretation of eulerian circuit on the complete graph, but the question is exactly to describe such a circuit, right? And I found it most convenient to interpret it as finite field elements... So I don't quite get your point. $\endgroup$ – WhatsUp Oct 24 '19 at 5:22
  • $\begingroup$ @JaapScherphuis Well... upon carefully reading the question, it does say "Can you place ...". So perhaps you mean a better answer should be "yes I can, because $7$ is an odd number"? $\endgroup$ – WhatsUp Oct 24 '19 at 5:26
  • $\begingroup$ I just meant that you can just start laying out a line of dominoes until you can't extend it any further. At that point it will be a circle (yes, because 7 is odd). It just seemed to be so oddly specific to construct the circles the way you did, making it seem as if there was a reason for it, as if only those circles will do. $\endgroup$ – Jaap Scherphuis Oct 24 '19 at 5:39

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