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 Attempt at a solution (probably too naive) by Charlie)

By using the same font as on a calculator it is possible to further increase the number of invertible numbers since 12 inverts to 21.

This increases Charlie's answer to 6(months) x 8(days) = 48

If we also allow the number format to be flexible as to whether or not there is a leading zero for single digit numbers then we can have 10 inverting to 01 and 20 inverting to 02.

So now the total increases to 6(months) x 10(days) = 60

On the number cube we have 01/10,02/20,6/9,12/21, if we make the remaining two faces 17 and 27 and make one of the faces on the month cube show "JU" for June. Then if we invert 17 it becomes L1 turning JU 17 into JUL 1 (similarly JU 27 becomes JUL 2) which adds another 2 onto the total. I'm not sure if this violates the rules as it puts the month across the two cubes - the last sentence of the problem "one month symbol...can be on 1 face of 1 cube" can be taken to mean 'no repeats' (in which case the solution is valid) or it can be taken as 'the month must appear on only 1 cube' (in which case these last two are not valid).

OK so the max I've got to so far is 62. Who's got the next bright idea?

Posted by fwaff on 2003-04-06 22:48:14