You have two six-sided cubes. Put one digit on each face using digits 0–9, and try to display every date from 01 to 31. For 01–09, a leading zero is required.
There is an important twist: under the strict literal rules, the puzzle is actually impossible. The famous classic version works only if 6 can be rotated and used as 9.
One valid arrangement is: Cube A = {0,1,2,3,4,5}, Cube B = {0,1,2,6/9,7,8}.
To show 01, 02, ..., 09, the zero must sometimes appear on the left while the other digit appears on the other cube. If only one cube had 0, the other cube would need digits 1–9 all by itself, which is impossible with only 6 faces.
A two-digit display uses one cube per digit. So to show 11, each cube must contain a 1.
Now each cube must already contain 0, 1, and 2. That uses 3 faces on each cube, or 6 faces total.
That is 7 more distinct digits, but there are only 6 remaining faces total across both cubes.
If 6 and 9 are treated as the same physical face, then the remaining needed symbols become: {3, 4, 5, 6/9, 7, 8}, which is exactly 6 symbols. That fits perfectly into the 6 remaining faces.