I have a small wooden calendar that uses two six-sided dice to display day of month. One die carries numbers 0, 1, 2, 6, 7, 8 and the other carries 0, 1, 2, 3, 4, 5. The six of course doubles as the nine.
Knowing the list of permutations is 01 through 31, with just a little guessing it was easy to arrive at which numbers need to appear on which die to meet the goal. How would these numbers be determined using only math?
On
Your use of 'solve' suggests that the configuration you've laid out is the only one that works, but in fact almost the opposite is true; many different assignments of digits will give the correct answer. azimut's answer hits on the key piece of the puzzle: because the 11th and the 22nd are both valid days of the month, then 1 and 2 have to appear on both dice. Because 0 can be followed by any other number, 0 must also appear on both dice (if 0 only appeared on one die, then all of the numbers 1-9 would have to appear on the other, and that's more than a d6 can hold even with the 6-for-9 swap).
So we already know that our two dice are 012ABC and 012XYZ, where ABCXYZ is some combination of the six digits 3-8 (with 9, again, being forced onto the 'flip 6'). But since 0 and 1 appear on both dice then it doesn't matter which die has the 3; we'll always be able to represent the 30th and the 31st. And since no date requires two numbers each of which is at least 3, it doesn't matter which way we lay out the digits among the two dice; we'll always be able to form every date by choosing whichever die has our 1s digit as the second die and using the appropriate one of 0, 1, 2 on the other die. This means that (012345, 012678) is a valid answer, but so is (012357, 012468) or (012347, 012568), etc - in fact, there are exactly $\dfrac{6!}{2\cdot3!\cdot3!} = 10$ different ways of assigning the digits 3-8 among the two dice, and all of them 'work' in that they can represent all the dates of the month.
Wow, didn't know that system!
Why does the system work?
For any number of a day involving only the digits 0,1 or 2 (these are day 01, 02, 03, 10, 11, 12, 20, 21, 22 of the month) there are two possiblities, since both dices have the right digits.
For any other number of a day there is only one possibility: There is exactly one digit which is not a 0, 1 or 2. If it is 3, 4 or 5, you have to use the second dice for it. If it is 6, 7, 8, 9 or 0, you have to use the first dice (not that 6 can also be used as 9).
Note that this system would still work if there are months with 32 days, but not for 33 days any more.
How can we find such a numbering system?
Now we forget about the numbering system and want to find a way put digits on two dice such that every day number 01, 02,..., 31 can be represented.
From the days 11 and 22 we see that both dice must have the digits 1 and 2. Now there are 8 blank sides left. On one dice there must be a digit 0. Since together with the other dice we can only display at most 7 combinations involving a 0 (using the trick that 6 and 9 are the same), but we need all the combinations 01,02,03,04,05,06,07,08,09, there must be a 0 also on the other dice. Now there are 6 blank sides left, which we have to use for the remaining digits 3,4,5,6,7,8 (9 is represented by 6). It is easy to see that we are free to distribute those 6 digits to the 2 dice in any way we like (since they only occur in combinations with 0, 1 or 2).