This article is from the Puzzles FAQ, by Chris Cole chris@questrel.questrel.com and Matthew Daly mwdaly@pobox.com with numerous contributions by others.

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.).

geometry/calendar.s

First note that there are *nineteen* different letters in the

month abbreviations (abcdef gjlmno prstuv y) so to get them all on the

eighteen faces of 3 cubes, you know right away you're going to have to

resort to trickery.

So I wrote them all down and looked at which ones could be

reversed to make another letter in the set. The only pair that jumped

out at me was the d/p pair. Now I knew that it was at least feasible,

as long as it wasn't necessary to duplicate any letters.

Then I scanned the abbreviations to find ones that had a lot of

common letters. The jan-jun-jul series looked like a good place to

start:

j a n u l

was a good beginning but I realized

right away that I had no room for duplicate letters and the second cube

had both a and u so aug was going to be impossible. In fact I almost

posted that answer. Then I realized that if Martin Gardner wrote about

it, it must have a solution. :-) So I went back to the letter list.

I don't put tails on my u's so it didn't strike me the first time through that n and u could be combined. Cube 1 Cube 2 Cube 3 j a n/u n/u l would let me get away with putting the g on the first cube to get aug, so I did. j a n/u g n/u l (1) Now came the fun part. The a was placed so I had to work around it for the other months that had an a in them (mar, apr, may). m a r d/p y (2) Now the d/p was placed so I had to work around that for sep and dec. This one was easy since they shared an e as well. d/p e s c (3) Now the e was placed so feb had to be worked in. f e b (4) The two months left (oct, nov) were far more complex. Not only did they have two "set" letters (c, n/u), there were two possible n/u's to be set with. That's why I left them for last. o t c n/u v (5) So now I had five pieces to fit together, so that no set would have more than six letters in it. Trial and error provided: j a n/u a b e g n/u l or, c d/p g r s m alphabetically: f l j y c d/p n/u m o e v t s n/u r o f b v t y

Without some gimmick the days cannot be done. Because of the dates 11 and

22, there must be a 1 and a 2 on each cube. Thus there are 8 remaining spaces

for the 8 remaining numbers, and because of 30, we put 3 and 0 on different

cubes. I don't think the way you allocate the others matter. Now 6 numbers on

each cube can produce at most 36 distinct pairs, and we need 31 distinct pairs

to represent all possible dates. But since 3 each of {4,5,6,7,8,9} are on each

cube, there are at least 9 representable numbers which can't be dates.

Therefore there are at most 27 distinct numbers which are dates on the two

cubes, and it can't be done. In particular, not all of {04,05,06,07,08,09} can

be represented.

The gimmick solution would be to represent the numbers in a stylised format

(like say, on a digital clock or on a computer screen) such that the 6 can be

turned upside down to be a 9. Then you can have 012 on both cubes, and three

each of {3,4,5,6,7,8} on the other faces. Done.

Example: 012468 012357

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