Adler-n-Subtract.com

Questions & AnswersProductsAbout Mitch Adler

Question

Dear Mitch, 

You said you'd reveal the answer to how you can get the thirteen digits necessary to form each day of each month of the year, even though each cube only has six sides.  6 + 6 = 12 sides, not thirteen. So what's the way?   Is there some kind of flap that sticks to one of the cubes to give you the extra number?

Thanks, 

Todd "Smith"

Princeton, New Jersey

Answer

Dear Todd "Smith",

No Flaps, sorry, but I would probably consider a method like that if there weren't a much, much easier and sensible method.

To review, the question gave some important hints.  Such as:  you need two 1's to form the 11th of each month, you need two 2's to form the 22nd of each month, and you need a zero on each cube so you can create 01, 02, 03, 04, 05, 06, 07, 08, and 09.  f you only had one zero, you would need a 1, a 2, a 3, a 4, a 5, a 6, a 7, an 8, and a 9 on the other cube -- which already would require more than the six sides that any cube has

So,

The way I did it the first time was I made two columns, one with each digit that one cube must have and the other column with each digit that the other cube must have.  Like this:

Cube #1:                       Cube #2:

     0                                   0

     1                                   1

     2                                   2 

That leaves three blank sides on one cube and here blank sides on the other.

Next, since the most days a month has is 31, there must be a 3 on block separate from  a block that has 0 and 1 (so you can make 30th, 31st.  That's easy, you can put the 3 on either cube.  (also don't forget that you'll need the 3 to make the third of every month, the 13th, and the 23rd).  This simply means that the three will have to be on a block separate from a block that has a 2 as well, and once again, that's easy because either cube will do that trick. 

So now:

Cube #1:        Cube #2:

0                      0

1                      1

2                      2

3

The 4 needs to be on a cube separate from 0, 1, and 2, because you will need to create 04, 14, and 24.

Either cube will work; I happened to put it on the first one.

So:

Cube #1:                    Cube #2:

0                                        0

1                                        1

2                                        2

3

Next, the 5, 6, 7, 8, and 9 also never appear in the position of first digit, because no month has forty-something days or fifty-something or sixty-something, seventy-something or eighty-something or ninety-something...

So, since cube #1 has five sides already taken, I positioned the 5 on the sixth side and then moved on to cube # 2 and put 6 on one blank side, 7 on another blank side, 8 on another blank side and - uh-oh, I believe we've now covered all twelve sides of the two cubes, yet have not found a place for the 9. 

In other words they look like this:

Cube #1:           Cube #2:

0                              0

1                              1

2                              2

3                              6

4                              7

5                              8

What about the 9??!!??!!

What will happen on the 9th of each month? (And the 19th and the 29th - most Februarys only have 28 days, that's true, but when it's time to leap...) 

ANSWER:  Since no month has 96 days or 69 days, the 9 and the 6 never have to show their faces on the same day; and that's a lucky thing, because when you need one of those two, and you don't see it, you turn the block with the other upside down!

A 6 turned upside down is a 9, and a 9 turned upside down is a 6. 

And that is that.

Next time, or soon, I'll tell you the one sentence answer to the question regarding getting seven days on six sides of the day-of-the-week block.

Until then,

Hope this helped,

Mitch