On the shelf in my study is a calendar.
The day of the month is shown by finding the correct digits on two cubes.
How is this done?
Each cube has 6 faces, so in total there are 12 digits displayed on them. Obviously, between the two cubes all 10 digits need to be represented (otherwise you'd get stuck somewhere in the first 10 days of a month). This only leaves the opportunity for 2 digits to be repeated between the 2 cubes. Certainly 1 and 2 need to be on both cubes (as each month has an 11th and a 22nd).
However, 0 also needs to be on both cubes (to be the leading digit for the first 9 days of the month, and 1-9 can't all be on the same cube).
So, we have an impossible problem - we need all of the digits 0-9, plus an extra 0, 1 and 2, making a total of 13 digits to fit onto 12 faces, so a calendar cannot be made using two cubes to show all of the necessary digits.
But I'm looking at one.
What's the answer?
Cheat.
Use a number font where the 6 can be turned upsidedown to make the 9 !!
