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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