My inventive friend Alex is always coming up with some new and interesting scheme. ‘What do you think of this idea for a desk calendar?’ he exclaimed, waving a sketch before my eyes (see the diagram). ‘Two cubes with a single digit on each face, and arranged to give any date in the month!’

Alex and his sketch disappeared out the door in a flurry of excitement, leaving me wondering whether there was, in fact, any way of bringing this latest plan to fruition. Obviously we need to be able to place digits on the cube faces in such a way that all the numbers 01, 02, 03, ... 29, 30, 31 can be obtained with the two cubes.

At first sight it seemed to me that there were not enough faces to go around, yet clearly Alex thought he had found a way. But how did he do it?

Hint 1

Try to create the dates in a systematic way, noting what digits are required.

Solution

It is clear that there must be a digit 1 and a digit 2 on each cube (to obtain 11 and 22). It is a little less obvious that there must be a digit 0 on each cube. For we must obtain the nine numbers 01, 02, 03, 04, 05, 06, 07, 08, 09. Since each cube has just six faces, this cannot be done using a single 0 on one of the cubes.

So far then we have this situation:

Cube 1: 0, 1, 2, _, _, _; Cube 2: 0, 1, 2, _, _, _.

But now we need to place the seven digits 3, 4, 5, 6, 7, 8, 9 on the remaining six faces! Fortunately we can use the symmetry of the 6 and the 9, obtaining the 9 by placing the 6 upside down. This observation shows that a solution may be possible, but we still need to check that there is a solution.

One possible way is to place 0, 1, 2, 3, 4, 5 on one cube, and 0, 1, 2, 6 (= 9), 7, 8 on the other. All the required date combinations can now be obtained.

Extensions

1. On what regular polyhedron could all the months of the year be displayed?

2. Are there solids other than cubes which could be used to display the dates of the month economically?

3. Using a cube with the digits 4, 5, 6, 7, 8 and 9 on its faces, and a cube with 0, 1, 2 , 3, 4 and 5 on its faces, we can obtain an unbroken sequence of years from (19)40 to (19)65. Can you allott digits to two cubes which will then give a longer unbroken sequence of years?

Hint 1

Solution

Extensions