From a set of dominos, select the 10 that show no number higher than 3. Arrange them in a triangle so that one of the numbers of every domino in the pattern is equal to the difference of the numbers on the domino above it (zero, if the domino above is a doublet). The example shown is not quite good enough. The two dominos beneath the 2-0 in the third row should each contain a 2, and the 3-3 does not. The 3-3 also fails to contain a 1, which is required by the 2-1 above it on the right. There are two closely-related solutions, not counting their reflections. Sadly the problem seems to have no solution for larger triangles containing more dominos, but the problem with 10 dominos is already a good puzzle. |
