Kalendar kwiz: Implementation

One way to solve this would be:

There are a total of 12 faces from 2 cubes.

Since, we require 11 and 22 as dates, we need both 1 and 2 in either cubes. Once 1 and 2 are  imprinted on both the cubes, we are left with 12 - 4 = 8 faces to write the remaining digits. 

Let's start with 0, then. If we imprint 0 on one of the blank faces of the left cube, we need to have 3, 4, 5, 6, 7, 8 and 9 on the right cube to have dates 01, 02, 03, 04....09. But, 3 to 9 need another 7 faces. But we have only 4 faces left(after imprinting 1 and 2 on the right cube). So, we need to imprint 0 on both the cubes. This leaves us with total 6 faces: 0,1,2 digits having been already imprinted on each of the cubes.

Now, we have remaining 3, 4, 5, 6, 7, 8 and 9 numbers left to write on the cube faces to be enable us to display all possible dates. But these are seven digits as  opposed to total six remaining faces. 

Now, we do a sleight of hand. We use 6 and 9 interchangeably by rotating the cube 180 degrees on the horizontal axis, whenever required.

Then, we have the following digits imprinted on the cubes to cater for all possible dates:

Cube one: 0, 1, 2, 3, 4, 5

Cube two: 0, 1, 2, 6, 7, 8

With this configuration all dates (from 01 to 31) can be displayed. We have specifically named the cubes Cube 1 and Cube 2, (and not Left Cube and Right Cube) because they can interchange their positions from left to right. For example, when the zero of Cube two is required on the left, Cube 2 is placed on the left and Cube 1 on the right.

Trivia: At eleven past one today, we have in the Hour Minute Month Day format: 1:11::11:1.