We received a lot of entries for this month’s puzzle, but only a few achieved the minimum possible sum of 210.
A few entries this month noted that the constraints in the problem can be shown to imply a lower-bound sum of 210. That being said, finding an arrangement turned out to be quite difficult. To the left is one of the 7 possible configurations that achieve this minimum sum. (And the only one to use ‘1, 3, 5, 15’ in the tetromino; the other 6 all use ‘1, 3, 4, 12’.)
Congrats to this month’s solvers!