# Problem 90: Cube digit pairs

![graphic](img090.gif)

Each of the six faces on a cube has a different digit (0 to 9) written
on it; the same is done to a second cube. By placing the two cubes
side-by-side in different positions we can form a variety of 2-digit
numbers. For example, the square number 64 could be formed: In fact, by
carefully choosing the digits on both cubes it is possible to display
all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49,
64, and 81. For example, one way this can be achieved is by placing {0,
5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.
However, for this problem we shall allow the 6 or 9 to be turned
upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3,
4, 6, 7} allows for all nine square numbers to be displayed; otherwise
it would be impossible to obtain 09. In determining a distinct
arrangement we are interested in the digits on each cube, not the order.
{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5} {1, 2, 3, 4, 5,
6} is distinct from {1, 2, 3, 4, 5, 9} But because we are allowing 6 and
9 to be reversed, the two distinct sets in the last example both
represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of
forming 2-digit numbers. How many distinct arrangements of the two cubes
allow for all of the square numbers to be displayed?