This is the last of this week’s domino puzzles. As before, take one full set of dominoes and remove the tiles with blanks, leaving you with 21 tiles.

The diagram below shows 8 rectangles placed at the vertices of an octagonal wheel. Each rectangle is to have one domino tile placed inside it. Every tile consists of two squares, each with some pips inside it. The number in the centre of the diagram is the product of the 4 squares taken from the 2 dominoes that are diametrically opposite each other. All four such products must equal the same total and be formed from different tiles from your set of dominoes. Next, orient each tile so that the square with the smallest number of pips is closest to the centre of the diagram, thereby forming a kind of “inner circle”. Calculate the sum of these eight squares, the “inner sum”.

The aim is to find the largest possible “inner sum” of the lowest value squares of the 8 dominoes placed so that the product of the squares of diametrically opposite dominoes is constant.

As an example, the diagram shows two domino tiles placed diametrically opposite each other. The product of their 4 squares is 5x2x3x4 = 120. The partial “inner sum” is 2+3 = 5. If this were a solution, the aim would be to complete the diagram with 6 more dominoes whose products (as defined above) are all 120, then to add the squares in the “inner circle”. Finally, you would need to establish that your final answer is, indeed, the maximum possible.

Good Luck!

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