A different number is placed at each vertex of a cube.
Each edge is given the greatest common divisor of its endpoints.
Can the sum of the edge numbers be equal to the sum of the vertex numbers?
Provide sufficient reason for your assertion.
This started as solution (I thought I had)
but and then devolved into a series of questions and ideas...
Exploring the extremes, we see that making all vertices
(all different) 1 or a prime would lead to a sum of
edges E=12. The minimum sum of vertices in this case
would be V = 1+2+3+5+7+11+13+17 = 57 and V-E = 45.
A much smaller difference may be reached by reducing
the vertices to 1,2,...,8. Optimally placed around the
cube, e.g. the bottom face: 1,2,4,5 and the top
face: 3,6,8,7 with 3 arranged above 1, 6 above 2,
etc, gives values: V=36, E=20, and a difference
of 16. These eight numbers may be oriented in
several different ways to give the same final
result.
Another improvement (bringing V and E
closer) may be reached by optimally
arranging some subset of 1-12 with examples shown:
lower upper
square square V E V-E
------------------------------------
1 2 3 5 4 8 6 10 39 25 14
1 2 3 5 4 8 6 12 41 27 14
1 5 2 4 3 9 6 12 42 28 14
1 2 3 5 4 12 6 10 43 29 14
1 2 3 9 4 8 6 12 45 31 14
There are many different arrangements
possible for the 8 vertex numbers for
each V, E pair.
14 is almost as close as I could make V and E get.
Raising any edge value from 1 to 2 doubles
one of the vertex values - e.g., along one
edge with end point v=2 and v=3, when
(v, e, v) is changed from (2,1,3) to (2,2,6),
we have raised E by 1 and V by 3, and made things
worse
So, placing larger numbers at the vertices will
only increase the gap between V and E.
(Except see below!) Perhaps they cannot be equal.
Later, to prove my point, I made
a table of a complete searches of numbers,
each time using a top limit on the max vertice
value.
Max Max Vertex Edge Least
Vert GCD Sum Sum Diff
--------------------------
8 4 36 20 16
9 4 38 22 16
10 5 39 25 14
11 5 42 22 20
12 6 41 27 14
13 6 46 24 22
14 7 45 27 18
15 6 48 32 16
16 8 52 37 15
17 6 50 24 26
18 9 55 39 16
19 6 52 24 28
20 10 51 33 18
21 7 56 34 22
22 11 57 31 26
23 6 56 24 32
24 12 60 47 13 *** BEST!!!
25 5 54 24 30
26 13 63 33 30
----
The best so far is the arrangement with 23. (!)
Eg. Bottom: 24 3 1 8 Top: 12 6 2 4
V=60 E=47
The fact that 24 has so many divisors is the key.
So, now I wonder now if there is a solution. I played with
having the vertices be 1,2,4,8,... since the sum of
2^i up to n is (2^n -1), and the extra two edges might
be used to add the extra 1. Or, if I left out the vertex
4 I could use a vertex "3" to add 1 to E and -1 to V.
This did not work, but it lead me to the question: what
really happens on the 3 edges reaching a 2^i
vertex? - if there is a 2^j opposed vertex, the
lesser of i and j wins the edge value. So, how to configure
vertices to sum to (2^n -1)? E.g. How to in0sure the each
2^i prevails once on an edge? I played but The best I
could reach was V-E=22.
Likewise, I explored the "perfect numbers" 496 and 8128,
whose factors sum to the whole. I could not het these to
work either - for 8128 I reached V-E=46. It made me wonder -
are there numbers whose factors sum to _more_ than the
number?
Maybe there are "close to perfect" numbers whose factor
sums exceed the number slightly that would be useful. This
is a well worn area of analysis but I am not conversant.
In conclusion, I don't have a conclusion - I don't know if
V-E = 13 is the minimum, or if V=E is possible.
--------------------
Side note: If the rules were relaxed and the vertex
numbers did not have to be different, there would
be a solution with all vertices=1 except a
pair of 3's diametrically opposed across the
main body diagonal. E=V=12.
Edited on March 4, 2024, 8:06 pm
soln
