Two-cube calendar

Digits 1 and 2 need to be placed on both cubes to allow numbers 11 and 22. That leaves us with 4 sides of each cube (total of 8) for another 8 digits. However, digit 0 needs to be combined with all other digits, so it also needs to be placed on both cubes. That means we need to place remaining 7 digits (from 3 to 9) on the remaining 6 sides of cubes. The solution is possible because digit 6 looks like inverted 9.

Therefore, the solution of the problem is:

${\displaystyle {\begin{cases}C_{1}:=\{0,1,2,3,4,5\}\\C_{2}:=\{0,1,2,{\tfrac {6}{9}},7,8\}\\\end{cases}}}$

If the problem is based on another given set of visible digits, the last three digits of each cube could be shuffled between the cubes.

A variation with three cubes providing English abbreviations for the twelve months is discussed in a Scientific American column in December 1977.[6] One solution of this variation allows displaying the first three letters of any month and relies on the fact that lower-case letters u and n and also p and d are inverses of each other.[7]

${\displaystyle {\begin{cases}C_{1}:=\{e,g,j,o,r,y\}\\C_{2}:=\{a,c,f,{\tfrac {n}{u}},s,v\}\\C_{3}:=\{b,{\tfrac {d}{p}},l,m,{\tfrac {u}{n}},t\}\\\end{cases}}}$

Polish 3-letter month abbreviations (informal but commonly used for date rubber stamps - sty, lut, mar, kwi, maj, cze, lip, sie, wrz, paź, lis, gru) are also feasible, both in lower and upper case:

${\displaystyle {\begin{cases}C_{1}:=\{a,e,g,l,w,y\}\\C_{2}:=\{c,i,j,r,t,{\acute {z}}\}\\C_{3}:=\{k,m,p,s,u,z\}\\\end{cases}}}$

Using four cubes for two-digit day number from 01 to 31 and two-digit month number from 01 to 12, and assuming that digits 6 and 9 are indistinguishable, it is possible to represent all days of the year. One possible solution is:

${\displaystyle {\begin{cases}C_{1}:=\{0,1,X,3,4,5\}\\C_{2}:=\{0,1,2,3,4,5\}\\C_{3}:=\{0,1,2,{\tfrac {6}{9}},7,8\}\\C_{4}:=\{0,1,2,{\tfrac {6}{9}},7,8\}\\\end{cases}}}$

The ${\displaystyle X}$ could be any digit. The last three digits of each cube could be shuffled between the cubes such that each digit from 3 to 9 is placed on at least two different cubes.

With assumption that 6 and 9 are distinguishable characters, it is impossible to represent all days of the year because the necessary number of faces would be 25 and four cubes have only 24 faces. However, it is possible to represent almost all days in the year. There is a family of the best solutions which excludes only one day, namely Nov 11, e.g.:

${\displaystyle {\begin{cases}C_{1}:=\{0,2,3,4,5,6\}\\C_{2}:=\{0,1,3,4,5,6\}\\C_{3}:=\{0,1,2,7,8,9\}\\C_{4}:=\{0,1,2,7,8,9\}\\\end{cases}}}$

The last four digits of each cube could be shuffled between the cubes such that no cube has two identical faces (especially the 2's).[8]

• Sicherman dice

• Martin Gardner (1992). Mathematical Circus. Washington, DC: MAA. pp. 186, 196–197. ISBN 0-88385-506-2. LCCN 92-060996.
• Ian Stewart (2010). "Perpetual Calendar". Professor Stewart's Cabinet of Mathematical Curiosities. Profile Books. pp. 35, 260. ISBN 1847651283.

• Jenny Murray. "The Colossal Book of Short Puzzles and Problems: Review by Jenny Murray". Association of Teachers of Mathematics. Archived from the original on 2015-05-09.