Solution to Enigma No. 1592

Answer: 0, 24, 27, 512 and 576

The winner is John Blankenbaker of Dallas, Texas, US

Worked answer

For the products to be different only one can be 0 so the first row has all six 0s.

Then, to accommodate the 0-1, 0-2, 0-3 and 0-4 the next row will include a 1, 2, 3 and 4.

The second row’s product cannot be odd or a power of 2.

Nor can it be a square, for that would need a 3 and 2 in its remaining places needing the products to be 0, 144, >144, >144, >144 – and we don’t have that many spots.

So, of the remaining three rows, one has an odd product (O) (so only uses 1s and 3s), just one is a power of 2 (P) (so uses only 1s, 2s and 4s) and the other (S) is a perfect square and not a power of 2 (so it uses at least one 3).

We need at least one 3 in S and hence (to make it a square) at least two 3s in S. So O (to have a product >24) must include {1,1,1,3,3,3} and S must include {3,3}. To keep the second row’s product between 0 and 27 we must have two 1s in its remaining places.

So the first row is {0,0,0,0,0,0}, the second is {1,1,1,2,3,4}, the third is {1,1,1,3,3,3} and that leaves {2,2,2,2,2,3,3,4,4,4,4,4} for the last two rows (which are S and P).

S/P must consist of {2,2,2,2,3,3}/{2,4,4,4,4,4} or {2,2,3,3,4,4}/{2,2,2,4,4,4} or {3,3,4,4,4,4}/{2,2,2,2,2,4} (with the lower of the two products underlined and thus forming the fourth row).

To fit the 1-2 and 1-4 in the layout the fourth row must include a 2 and a 4, ruling out the first of those three possibilities. The bottom row’s product is less than 1000, ruling out the third of those possibilities. Hence the fourth row is {2,2,2,4,4,4} and the bottom row is {2,2,3,3,4,4}. This gives products 0, 24, 27, 512 and 576.

To check that a layout is possible, see Figure 1 (top right).