## Wednesday, May 19, 2010

### Cubic Calendar

From http://www.folj.com/puzzles/
A corporate businessman has two cubes on his office desk. Every day he arranges both cubes so that the front faces show the current day of the month.

What numbers are on the faces of the cubes to allow this?

Note: You can't represent the day "7" with a single cube with a side that says 7 on it. You have to use both cubes all the time. So the 7th day would be "07".

Solution thanks to Rob Strong:
you need two 1's, and two 2's; you only need one 0, 3, 4, 5, 6, 7, 8, 9.

That's twelve digits you need, which is very convenient.

However, you need to be able to display nine different days beginning with 0, ten different days beginning with 1, and ten beginning with 2. Because you can't fit ten numerals on a single cube, you'll need to have 0, 1, and 2 represented on both cubes. We knew that about 1 and 2 (since we'll need to display 11 and 22), but now we have a problem, because two 0's makes thirteen numerals needed, and only twelve faces on which to fit them.
HOWEVER! We are saved by the rotational symmetry of 6 and 9, and the fact that February 96th only comes around once every ten billion years, far outside the scope of our businessman's lifespan.
Thus the dice read as follows:

First cube:
0
1
2
3
4
5

Second Cube
0
1
2
7
8

1. you need two 1's, and two 2's; you only need one 0, 3, 4, 5, 6, 7, 8, 9.

That's twelve digits you need, which is very convenient.

However, you need to be able to display nine different days beginning with 0, ten different days beginning with 1, and ten beginning with 2. Because you can't fit ten numerals on a single cube, you'll need to have 0, 1, and 2 represented on both cubes. We knew that about 1 and 2 (since we'll need to display 11 and 22), but now we have a problem, because two 0's makes thirteen numerals needed, and only twelve faces on which to fit them.

HOWEVER! We are saved by the rotational symmetry of 6 and 9, and the fact that February 96th only comes around once every ten billion years, far outside the scope of our businessman's lifespan.

Thus the dice read as follows:

First cube:
0
1
2
3
4
5

Second Cube
0
1
2
7
8

2. "and the fact that February 96th only comes around once every ten billion years, far outside the scope of our businessman's lifespan."

I like it

3. Oh... I was thinking the businessman had a geeky side as I do.

If so, he would just use two cubes (0-5) and represent the day of month in base 6 which counts to 35-decimal.

http://en.wikipedia.org/wiki/Senary#Finger_counting

Since most months of most years have at most 31 days, we have five possibilities left (32,33,34,35,00) so we may represent Feb. 96'th and four other rarely occurring dates!

1. Oops getting my decimal and senary mixed. The possibilities left should instead be: 52,53,54,55,00 ...which in decimal is what I wrote above (32,33,34,35,0). Thanks! :)