Ken's POTW
Sums on a Cube
ab
/: /
/ : / 
dc 
 h....g
 /  /
/ /
ef

 Can you place the numbers 18 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 18 at the corners of a cube, such that each
 face has the same sum (of its four adjacent corners)?
Yes.
14
/: /
/ : / 
85 
 7....6
 /  /
/ /
23
Face sum = 18
 In how many ways can this be done?
Ignoring any reductions due to symmetry, there are 144 eqisum faced
cubes out of a total of 40320 cubes for (ah) = (18).
[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.
12
/: /
/ : / 
43 
 6....8
 /  /
/ /
57
1/4: Face sums = 10 26 20 16 19 17
Ignoring any reductions due to symmetry, there are 13584 uniquesum faced
cubes out of a total of 40320 cube for (ah) = (18).
 Consecutive sums?
For (ah) = (18) it is not possible to find any uniquesum faced cube
where the the sums are consecutive.
For (ah) = any 8 of (19), there are 2880 uniquesum faced cubes where
the sums are consecutive out of a total of 362880. Example:
12
/: /
/ : / 
95 
 6....7
 /  /
/ /
34
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 (ah) = (18), there are 9744 uniquesum cornered cubes out of a
total of 40320. Example:
12
/: /
/ : / 
43 
 6....8
 /  /
/ /
57
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 (05),
so that the vertex sums are consecutive (411):
front: 0
back: 2
left: 4
right: 3
top: 1
bottom: 5

Sums:
76
/ /
54 
1110
/ /
98

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