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.

(In reply to re: Attempt at a solution - more combinations by fwaff)

Ok now this is getting tenuous so bear with me....

Instead of the 'calculator font' where 2 is represented with 5 straight lines, draw it with 3 straight lines such that it looks like Z. It's still horizontally symmetrical so Z0/0Z and 1Z/Z1 remain valid.

Now take the ZO face and turn it 90deg clockwise so that it shows NO reading vertically downwards. Now we have a month on the 'day' cube, so what we need are some days on the 'month' cube. The only two that I can see are S (for September) looks like 5 (giving NO 5 - bonfire night) and if you take the letter M (for March or May) and rotate it 90deg clockwise it looks like 3 (for NO 3).

So far the cubes have:

Cube 1: 01(10), 02(20,NO), 6(9), 12(21), 17(L1) and 27(L2)
Cube 2: JU, S(5), M(3), ???, ***, @@@

...which give 64 combinations. Only 302 left to find now!!!

Posted by fwaff on 2003-04-07 03:34:09