## Ken's POTW

Sums on a Cube
 ``` a-----b /: /| / : / | d-----c | | h..|..g | / | / |/ |/ e-----f``` Can you place the numbers 1-8 at the corners of a cube, such that each face has the same sum (of its four adjacent corners)? In how many ways can this be done? Can you place different numbers (preferably consecutive) at the corners of a cube such that the six faces have different sums? Consecutive sums? Can you place different numbers (preferably consecutive) on the sides of a cube such that the eight corners have different sums (of their three adjacent faces)? Consecutive sums?
Source: Original, but I wouldn't be surprised to find them somewhere else. Please feel free to let me know where, or let me know of a related puzzle.
Solutions were received from Art Hagar and Kirk Bresniker. Kirk's solutions follow.
```| 1.Can you place the numbers 1-8 at the corners of a cube, such that each
| face has the same sum (of its four adjacent corners)?

Yes.

1-----4
/:    /|
/ :   / |
8-----5  |
|  7..|..6
| /   | /
|/    |/
2-----3
Face sum = 18

| In how many ways can this be done?

Ignoring any reductions due to symmetry, there are 144 eqi-sum faced
cubes out of a total of 40320 cubes for (a-h) = (1-8).

[After reducing the results, it turns out there are only three
unique cubes. -KD]

| 2.Can you place different numbers (preferably consecutive) at the
| corners of a cube such that the six faces have different sums?

Yes.

1-----2
/:    /|
/ :   / |
4-----3  |
|  6..|..8
| /   | /
|/    |/
5-----7
1/4: Face sums = 10 26 20 16 19 17

Ignoring any reductions due to symmetry, there are 13584 unique-sum faced
cubes out of a total of 40320 cube for (a-h) = (1-8).

| Consecutive sums?

For (a-h) = (1-8) it is not possible to find any unique-sum faced cube
where the the sums are consecutive.

For (a-h) = any 8 of (1-9), there are 2880 unique-sum faced cubes where
the sums are consecutive out of a total of 362880. Example:

1-----2
/:    /|
/ :   / |
9-----5  |
|  6..|..7
| /   | /
|/    |/
3-----4
1/2043: Face sums = 21 20 19 18 17 16

I didn't find any consecutively numbered cubes which had consecutive
sum, but I didn't explore that many.
[Consecutive sums can't be found using consecutive numbers. - KD]

| 3.Can you place different numbers (preferably consecutive) on the sides
| of a cube such that the eight corners have different sums (of their
| three adjacent faces)?

For (a-h) = (1-8), there are 9744 unique-sum cornered cubes out of a
total of 40320. Example:
1-----2
/:    /|
/ :   / |
4-----3  |
|  6..|..8
| /   | /
|/    |/
5-----7
1/4: face sums = 10 26 20 16 19 17
1/4: corner sums = 65 61 49 47 45 43 63 59

[Kirk actually solved problem 2 and 3 with this example, but I had
meant for problem 3 to be solved with the smallest sums, without needing
to solve problem 2 with the same cube. - KD]

| Consecutive sums?

```
Nick Baxter sent this solution 1/28/97:

Here is a consecutive numbering for the faces (0-5), so that the vertex sums are consecutive (4-11):
 ```front: 0 back: 2 left: 4 right: 3 top: 1 bottom: 5 ``` ```Sums: 7---6 /| /| 5---4 | |11-|-10 |/ |/ 9---8 ```

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