# Turn the dials until each of the 12 columns add up to 42

I received the wooden puzzle below for Christmas. After playing with it for a while I decided I needed to actually write something down if I was going to solve it. Eventually I just wrote an algorithm that went through the 20736 possible positions of the 4 dials and found the only answer. The problem is I feel like I cheated/missed something.

Could I have done better than brute force?

Here is the puzzle.

On the rear is the text:

Solving the Greek Computer. Turn the dials until each of the 12 columns add up to 42

And here are the tables I created of the puzzle so you have all the numbers (0 always represents a hole in that dial):

3   0   6   0   10  0   7   0   15  0   8   0
0   0   0   0   0   0   0   0   0   0   0   0    Dial 1 (top)
0   0   0   0   0   0   0   0   0   0   0   0
0   0   0   0   0   0   0   0   0   0   0   0

6   17  7   3   0   6   0   11  11  6   11  0
12  0   4   0   7   15  0   0   14  0   9   0    Dial 2
0   0   0   0   0   0   0   0   0   0   0   0
0   0   0   0   0   0   0   0   0   0   0   0

9   13  9   7   13  21  17  4   5   0   7   8
21  6   15  4   9   18  11  26  14  1   12  0    Dial 3
5   0   10  0   8   0   22  0   16  0   9   0
0   0   0   0   0   0   0   0   0   0   0   0

11  0   8   0   16  2   7   0   9   0   7   14
14  12  3   8   9   0   9   20  12  3   6   0    Dial 4
9   0   17  19  3   12  3   26  6   0   2   13
6   0   10  0   10  0   1   0   9   0   12  0

11  11  14  11  14  11  14  14  11  14  11  14
4   5   6   7   8   9   10  11  12  13  14  15   Base (bottom)
4   4   6   6   3   3   14  14  21  21  7   9
8   3   4   12  2   5   10  7   16  8   9   8


And the solution:

(note I marked the answer so I can get to it easily)

• Welcome to Puzzling! Can you clarify if every zero in the text version is a "hole" in the physical puzzle (so you can see through to the layer below), or are there actual zeros anywhere?
– fljx
Jan 24 at 12:04
• Yes, all the zeros are holes that reveal a number below. All the numbers start from 1. Jan 24 at 12:34
• You are relying on the picture - do you have 5 wheels that are freely to rotate?
– Moti
Jan 25 at 5:24
• I would consider the base static (the last table) then the other 4 dials freely rotate above it. That gives me 12^4 = 20736 positions of the 4 dials above the base. Jan 25 at 10:12
• I have one of these and FWIW in the last two rows, next-to-last column, the 7 and 9 are transposed on mine but there is still only one solution that I found (they are unused). I haven't made any progress on a logical solution. Feb 3 at 15:33