I ran into a fun little day-indicator that is made from two cubes with Arabic numerals (0, 1, …). I think this is fun because it seems impossible to create all 31 numbers (01, …, 31) with only two cubes.

Let’s figure out how to construct the cubes. Each face can have one number (0, 1, …, 9), and we have two cubes.

First we need to figure out what number combinations we need. A few observations:

• With ‘1’ and ‘3’ on separate cubes we can make ‘13’ and ‘31’.
• We always need to use two cubes. The first of the month is ‘01’.
• We need some numbers on both cubes (to make ‘11’, ‘22’).

Let’s try and construct the minimal set greedily:

We need ‘0’, on one cube, and leave the choice for the other faces open:

${ 0, ., ., ., ., . }$

${., ., ., ., ., .}$

Now, to make ‘01’, ‘02’, …, we need to put the ‘1’, ‘2’, ‘3’, … on the other cube. Unfortunately this only allows us to make six out of nine ‘0x’ numbers, by the pigeon-hole principle.

To be able to construct the remaining numbers we need to put ‘0’ on the other cube as well. We now have:

${0, ., ., ., ., .}$

${0, ., ., ., ., .}$

We already know we need ‘1’ and ‘2’ on both cubes as well (to make ‘11’, ‘22’):

${0, 1, 2, ., ., .}$

${0, 1, 2, ., ., .}$

Let’s try and greedily fill in the remaining numbers and see where we’re at:

${0, 1, 2, 3, 4, 5}$

${0, 1, 2, 6, 7, 8}$

Great, but we can’t fit the ‘9’. In fact, there’s no way to fit all numbers we need, as we have seven numbers beyond ${0,1,2}$ and only six faces to put them on.

What if I tell you we actually have everything we need?

Do you see the trick?

If not, try turning this page upside down (I hope you don’t have a big monitor). Or look carefully at the photo of the cubes.

I think this is a pretty neat example of “out of the box thinking”.