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:
 All six faces show four different numbers but each of the numbers 1, 2, 3, 4, 5, and 6 appear four times.
 The numbers on the front can be arranged to make a fourdigit square number, and those on the back can form a different fourdigit square.
 The sum of the numbers showing on the top is the same total as those on the base.
 The numbers showing at corner F are 2, 3, and 6.
 The numbers seen on the front are consecutive, but not necessarily in order.
 Three odd numbers form corner B, and the product of all the numbers on the top is 180.
 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.
 The numbers on the right side do not include a 6, and only one 5 appears in the top half of the cube.
 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
EF
 2  6 
+
back left  3  5  right
FEA+BF
 2  6  4  2  6  3  1  3 
+++++++
 1  4  1  3  5  4  5  2 
GHD+CG
 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/diceplay/DiceStandard.htm), when you place the 1 on top and the 2 in front, the 3 will be on the right. These are called righthanded dice. Lefthanded 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 oppositesidesaddto7 rule determines the other numbers' positions accordingly.
A righthanded 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: 236 clockwise  righthanded
B: 135 clockwise  lefthanded
C: 145 clockwise  righthanded
D: 356 clockwise  righthanded
E: 246 clockwise  lefthanded
F: 623 clockwise  righthanded
G: 142 clockwise  lefthanded
H: 154 clockwise  lefthanded
So you'd need four western dice and four Chinese dice.

Posted by Charlie
on 20041222 19:13:04 