How many different ways are there to assign the ten decimal digits (0–9) to each face of two cubes to ensure that we can arrange both cubes on any day such that the front faces of the cubes show the current day of the month?

For example, on February 9 the cubes would be placed side by side such that the front face of the cube on the left side shows 0 and that of the one on the right side shows 9.

Two ways of assigning the digits to the faces of the cubes are considered different if and only if it is not possible to get one assignment from the other by performing one or more of the following operrations:

  1. Rotating (reorienting) the digits with respect to the faces they belong to.
  2. Rotating the cubes.
  3. Swapping the cubes.

1 comment

Alex Yeilding solved this puzzle:

This only works with one face used for both the 6 and 9.

Both cubes need to have 0, 1 and 2. The remaining six digits can be assigned to the cubes in C(6, 3) / 2 = 10 ways.

There are 5 choices for a number to go on the face opposite the zero. Then put the lowest remaining number on an open face, and there are 3 choices for a number to go opposite that one.

There are 2 choices for how to arrange the final two digits.

So now we are up to 10 · 302 = 9000 unique ways to assign digits to the faces.


This is an original cotpi puzzle inspired by a puzzle from folklore:

Further reading

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