Answer to the January 19, 2004 Problem of the Week

The Calendar Cube Problem   by Martin Gardner and Michael J. Maggio

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If you haven't purchased your Elvis Calendar by now, you're probably out of luck!

So you will need to build a calendar from two sets of cubes.

On the first set, spell the months with a letter on each face of three cubes. Use lowercase three-letter English abbreviations for the names of all twelve months (e.g., "jan", "feb", "mar").

On the second set, number the days with a digit on each face of two cubes (e.g., "01", "02", etc.).

Show what letters and digits need to be put on the five cubes so that you will be able to display all 366 days correctly!

 

Solution to the Problem:

Cubes for the months:

a, e, l, t, u/n, v
f, g, j, m, o, p/d
b, c, n/u, r, s, y

Cubes for the dates:

0, 1, 2, 3, 7, 8
0, 1, 2, 4, 5, 6/9

You have 12 months, with 19 unique letters.
However, since the letters are lowercase, you have the freedom to "flip" the cube over to get the other letter, such as flipping "p" to get "d" and "n" to get "u".

Doing this, you find that you can only get rid of one letter, either the "p" or the "d". You cannot do this with the "n" and "u" since both of them are needed for June (jun).

With that, each month has one vowel -- even August, since you can easily substitute the "n" for the "u". After this step it's mostly logic, since you now have 18 letters.

One of the easiest ways, then, is to start out with the "a" since the "a" cannot be on the same cubes with the most letters (j n m r p y u g).

With the dates, 6 and 9 are reversible and you never need both of them at one time.

Nearly everyone sent in a different set of letters that worked, and James Alarie sent in three different valid sets.