Professor Buric likes to set his students logic problems using eight normal dice in a 2󫎾 cube as shown here. Let the six faces of the cube be referred to as: ABFE - TOP ABCD - FRONT DCGH - BASE EFGH - BACK AEHD - LEFT BFGC - RIGHT

Each of the six faces shows four numbers and three faces of one die are seen at each of the eight corners of the cube. The Professor then makes several statements regarding the viewable numbers and challenges the students to place all twenty four visible numbers in their correct positions in 40 minutes or less.

On one occasion the following clues were given:

1. All six faces show four different numbers but each of the numbers 1, 2, 3, 4, 5, and 6 appear four times.
2. The numbers on the front can be arranged to make a four-digit square number, and those on the back can form a different four-digit square.
3. The sum of the numbers showing on the top is the same total as those on the base.
4. The numbers showing at corner F are 2, 3, and 6.
5. The numbers seen on the front are consecutive, but not necessarily in order.
6. Three odd numbers form corner B, and the product of all the numbers on the top is 180.
7. At corner G, the sum of the numbers is only 7, while at corner E, only even numbers are seen, and each of these is either the largest or smallest number seen on its face.
8. The numbers on the right side do not include a 6, and only one 5 appears in the top half of the cube.
9. At corner C, one of the numbers equals the sum of the other two.

Knowing that the dice need not be identical, as the only rule governing the positions of the numbers on these dice is that opposite faces add up to seven, can you graduate from the Professor's class?

(In reply to Solution : Method and Spoiler by DJ)

DJ did the puzzle much better than I did.

Here's another diagram of DJ's solution:
                     top
                  E-------F
                  | 2 | 6 |
                  |---+---|
     back    left | 3 | 5 | right
  F-------E-------A---+---B-------F
  | 2 | 6 | 4 | 2 | 6 | 3 | 1 | 3 |
  |---+---+---+---+---+---+---+---|
  | 1 | 4 | 1 | 3 | 5 | 4 | 5 | 2 |
  G-------H-------D---+---C-------G
                  | 6 | 1 |
                  |---+---|
                  | 5 | 4 |
                  H---+---G
                   bottom

                  
Note that if you try to construct this with real dice, you will find that some of the dice are mirror images (i.e., backwards) of the others.  Standard dice in most of the world (the exception being China, according to http://homepage.ntlworld.com/dice-play/DiceStandard.htm), when you place the 1 on top and the 2 in front, the 3 will be on the right.  These are called right-handed dice.  Left-handed dice, presumably used in China, are the other way around: if you put the 1 on top and 2 in front, the 3 will be on the left.  Then the opposite-sides-add-to-7 rule determines the other numbers' positions accordingly.

A right-handed die:

    5
  4 1 3 6
    2

   
also viewable as (useful in judging CW from CCW when 6 is involved with 2 and 4):

    5
  4 1 3
    2
    6

In the solution, the dice show:

A: 2-3-6 clockwise -- right-handed
B: 1-3-5 clockwise -- left-handed
C: 1-4-5 clockwise -- right-handed
D: 3-5-6 clockwise -- right-handed
E: 2-4-6 clockwise -- left-handed
F: 6-2-3 clockwise -- right-handed
G: 1-4-2 clockwise -- left-handed
H: 1-5-4 clockwise -- left-handed
                   
So you'd need four western dice and four Chinese dice. 

Posted by Charlie on 2004-12-22 19:13:04