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Calendar Cubes pt 3 (Posted on 2003-04-06) Difficulty: 4 of 5
In the problem Calendar cubes you figured out the maximum amount of days you could fit on two cubes by putting numbers on both cubes and using the faces of the cubes to combine and make more numbers.
In Calendar cubes pt 2 You figured out how many months you could fit on two cubes.
Now in calendar cubes pt 3 you must figure out how many days of the year you can fit on two cubes. E.G one cube says mar(for march) and another says 5. so you could make the date march 5 and that would count as one date.

To represent months you may use
a) the first letter of that month
b) the first and second letter of that month
c) the first three letters of that months. So for january you could use either j, ja or jan to represent that month.
Also no two letters or letter combinations can represent the same month. So j cannot stand for june and july, but you can have j stand for june and ju stand for july. Also note that one month symbol (lets say au for august) can be on 1 face of 1 cube.

See The Solution Submitted by Alan    
Rating: 2.3000 (10 votes)

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Some Thoughts re: What? | Comment 11 of 14 |
(In reply to What? by Alan)

Although it is not stated explicitly the implication is that each cube shows one full face. If we are allowed to display the cubes such that one or two faces are visible then other combinations appear. For example, if we have j, u, a, m, o, f on the faces of one cube then it can be positioned to show:

One face showing
j - july
a - april
n - november (turn the u upside down)
o - october
m - march
f - february

Two faces showing
ja - january
ju - june
au - august
ma - may

Similarly, having 1 and 2 on opposite faces of the other cube and the numbers 3,4,5,6 on the other faces it's possible to get:

One face
1,2,3,4,5,6,9

Two faces
13,14,15,16,19,23,24,25,26,29

Which gives a total of 17x10 = 170 combintions (not counting combinations using letters as numbers and vice versa eg the 5 as an S and the M as a 3, etc)

Conversely, if the cube holder is constructed in such a way that only half a face is visible (like the one Nick Reed's nan has - see his comment in the orginal Calendar Cubes problem) then it's easy to have all twelve months on one cube. Similarly make the day cube such that only one corner of a face is visible, then 24 days are easily possible.

This gives 12x24 = 288 combinations.

Alan: are either of these the 'genius concept' you had in mind?

  Posted by fwaff on 2003-04-10 22:52:35

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