A two-cube calendar seems impossible: too many dates and too few faces. But the days of the month do not demand all digits equally. Some digits must be duplicated, and one digit can hide another if it is turned upside down.
The Two-Cube Calendar
Riddle statement
A desk calendar shows the day of the month using two cubes. Each cube has one digit on each of its six faces.
By placing the two cubes side by side, it must be possible to display every date from 01 to 31. The 6 face may be turned upside down to represent a 9.
Give a valid numbering for the two cubes. It does not have to be unique.
Show solution
Solution
One valid numbering is:
First cube:
Second cube:
The 6 face is turned upside down to show 9.
The important constraints are these:
- both cubes need 0, because we must show 01, 02, ..., 09 by using the zero from either cube;
- both need 1 and 2, to form 11 and 22;
- a single face is enough for 6 and 9;
- the 3 only needs to pair with 0 and 1, to form 30 and 31.
The remaining digits can be distributed in more than one way. For example, swapping 4 and 7 between the cubes still gives a valid solution. That is why the riddle asks for one possible numbering, not the only one.
Answer: one solution is $0,1,2,3,4,5$ and $0,1,2,6,7,8$, using 6 also as 9.