-1

Using standard playing dice, in what way can you to stack up the dice to build a 1 X 1 X 6 high tower that do not have the same dice faces showing on all its vertical wall?

TSLF
  • 6,656
  • 13
  • 34

3 Answers3

4

The only constraints we have is that opposite faces have to sum to 7, and no number can appear twice at the same height of the wall. This gives us plenty of solutions, for example:

Face 1 of the wall has 1,2,3,4,5,6.
Face 2 of the wall has 2,1,6,5,4,3.
Face 3 of the wall has to have the numbers opposite face 1, so 6,5,4,3,2,1.
Face 4 of the wall has to have the numbers opposite face 2, so 5,6,1,2,3,4.

 6  3  1  4
 5  4  2  3
 4  5  3  2
 3  6  4  1
 2  1  5  6
 1  2  6  5
ffao
  • 21,733
  • 4
  • 68
  • 109
2

Okay, so assuming these dice are set up like standard 6 sided dice, this means that 1 is opposite 6 on the die, 2 is opposite 5, and 3 is opposite 4.

It doesn't really matter what order you stack the six dice in as long as there are no repeats, so let's assume we are stacking them so that our first side is in order 1, 2, 3, 4, 5, 6.

This necessarily gives us our third side (directly across from the first side) as

6, 5, 4, 3, 2, 1.

There are probably multiple answers to this, but my final dice stack looks like this, where each column represents a side of the tower

1  3  6  4
2  6  5  1
3  5  4  2
4  1  3  6
5  4  2  3
6  2  1  5
MMAdams
  • 3,662
  • 17
  • 35
2

Here is one way:

Place the bottom die with 1 down, the next with 2 down and so on, and turn each one to not match the ones below, by the time you get to the top there will be only one way you can complete your tower:

2 3 5 4
1 4 6 3
6 5 1 2
5 6 2 1
4 1 3 6
3 2 4 5

As it happens this is a variant of the now accepted answer - turn that upside down and then move the top two to the bottom.

Jonathan Allan
  • 21,150
  • 2
  • 58
  • 109