1.5 KiB
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:
+-------+ +-------+
/ /| / /|
+-------+ | +-------+ |
| | | | | |
| 6 | + | 4 | +
| |/ | |/
+-------+ +-------+
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?