Lots of great input, phrontister!
You must have a lot of boring meetings! ;-)
You are nearing the first challenge, I think.
What you have identified so far was very similar to what I did (determining the solutions are only 11,12 and 13; using excel to auto-sum the edges/corners; color-coding).
Although I hate to admit that I dont know how to attach images in these posts so I didnt provide the sample before.
Even if we dont get the real answer, suffice to say there are a lot of solutions (even though it does take some time to find them).
For the second challenge, I am interested to see if there is some minimal "seed" numbers that can guarantee the unique solution.
Maybe starting from the winning solution then "unfreezing" numbers until more than one solution can be created is one way to do it?
There are less than 890939317237438025973104 solutions which is a very large number. I just do not know how many less.
I am a little rusty, but I believe that each cube face can be arranged in 9! combinations, therefore there should be 9!^6 possible combinations across the entire cube.
This indeed is a very large number.
But since there are limitations to only summing to 11, 12 or 13 the number of combinations should be something much less.
Mathematically there should be solutions summing to 11, 12 and 13 - although I have done simulations with 11 and 12 my patience ran out trying to do the 13.
And I would guess there are many combinations of solutions.
Here is the challenging questions (which I dont know the answers) -
1. What are the number of solutions?
2. What are the minimal number of "fixed/seed" squares (squares which have already a number filled in and cannot be changed) that will guarantee only a single solution?
Not sure if this is an existing puzzle, my basic internet search has not found anything similar.
I came up with this puzzle while sitting in boring meeting at work!
Take a cube and divide each face into 3x3 squares.
On each cube face enter the numbers 1 - 9 in the squares without repeating such that the sum of all squares which share an edge or corner around the cube is equal.
Hint - there are 8 corners with 3 squares to be summed and 12 edges with 2 squares to be summed - all must be equal.