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:
"Two dices are marked with numbers from zero to nine on each face.
The two dices are meant to display all the dates in a month from 01 to 31.
Remember, single digit dates would have to be shown as 01, 02, 03 etc.
That is, one dice should show zero and the other 1 or 2 or 3 and so on.
The two dices can be interchanged i.e. placed in any order.
What would be the numbers written on the six faces of the two dices???"
I would like to know how to prove this situation is impossible if you don't count 6 and 9 as the same. Although I have a crude prose proof, I want to use just math and variables. How can I do this? Thanks