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Oh yeah, here is my written proof: The numbers 0-9 must be present at least once to represent the desired numbers 01-09. Therefore, 10 of the 12 faces of the two dice are taken. There must also be duplicates of the numbers 0-2 to represent the tens place in the dates. If 0-2 were present only once on one die, the other die could only represent 6 out of the 10 numbers need for 01-09, 10-19, and 20-29. To have duplicates of 0-2, there needs 3 more numbers on the faces of the die. The total number of faces needed is 10 + 3, 13. Because the dice only have 12 faces, the proposed situation is impossible.
ganesh posted this problem a few months ago: