In the arrangement of five dominoes shown in Figure 1, obeying the usual rule that adjacent squares have matching numbers, the two at the end total five spots, and so do the three in the middle. There are three other ways to achieve this, not counting left-right reflections as different. Can you find them?
This problem originally appeared in the Prisoner’s Dilemma in the 2024 Spring issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.
Show solution?Anthony Batiste and the Georgia Southern University problem group solved this dilemma. One can generate the solutions by assigning variables to each of the squares in turn from left to right, noting that every touching pair must be equal, and then finding integer solutions where the two required totals are equal to five. That analysis has quite a few cases, so we won’t give it in detail here, because Ian has already provided the information that there are exactly three other solutions. Therefore, it suffices to simply show the solutions, which we do in Figure 11.
