On my
request Frank Rubin proposed one puzzle for my pages:
It is possible to write a prime on each face of a cube so that the 3
faces surrounding each vertex sum to a prime. For example, if you
write 3 on a pair of opposite faces, and 2 on the other 4 faces, then
the 3 primes around each vertex sum to 7. Find the set of 6 distinct
primes with this property such that their sum is minimal.
The particular question posed by Rubin is not really hard to find even
without a computer program. A little harder problem is to find a solution of
minimal sum using six consecutive primes.
...
What if we ask both questions for those platonic solids who have an odd
number of faces contiguous to each vertex - tetrahedron, cube (or
hexahedron), dodecahedron and icosahedron?
(shown here thanks to the Rüdiger Appel's
page)
In order to organize the search of the answers to this puzzle we need to
use the planar projection of the platonic solids as follows:
| Tetrahedron |
Hexahedron
|
Dodecahedron |
Icosahedron |
| Faces contiguous to each
vertex |
| 3 |
3 |
3 |
5 |
 |
 |
|
|
Please download this Excel worksheet in order
to see the variable names that I propose for each face and each vertex, for
each platonic solid planar projection. Please use this proposal in order to
compare easily the distinct solutions gotten by different puzzlers.
Example.
For a tetrahedron (faces: a, b, c, d; vertexes: A, B, C, D)
1) Using distinct primes for the faces, the minimal sum is gotten with
the following values:
a = 5, b = 7, c = 17, d = 19, S = a+b+c+d = 48
A=a+b+c = 29, B=a+b+d=31, C=a+c+d=41, D=b+c+d=43
2) Using consecutive primes for the faces, the minimal sum is gotten
with the following values:
a = 19, b = 23, c = 29, d = 31, S = a+b+c+d = 102
A=a+b+c = 71, B=a+b+d=73, C=a+c+d=79, D=b+c+d=83
Questions:
1. Find a solution for the remaining three platonic solids
(cube, dodecahedron, icosahedron) such
that the sum of all the primes in the faces, S, is minimal: 1) using distinct primes
2) using
consecutive primes.
2.
Explain why no solution exist for a tetrahedron using distinct primes, one
of each being the prime '3'.
Contributions came from Torbjörn Alm and Ray Opao
Torbjörn Alm solved completely the question 1.
1.1) Distinct primes
| Cube |
Dodecahedron |
Icosahedron |
Sum = 72
Faces: a:3 b:5 c:13 d:17 e:11 f:23
Vertices:
A = e+a+b = 19
B = e+b+c = 29
C = e+c+d = 41
D = e+d+a = 31
E = f+a+b = 31
F = f+b+c = 41
G = f+c+d = 53
H = f+d+a = 43 |
Sum = 272
Faces: a:3 b:5 c:31 d:19 e:17 f:41 g:29 h:37 i:11 j:43 k:23 l:13
Vertices:
A = k+e+a = 43
B = k+a+b = 31
C = k+b+c = 59
D = k+c+d = 73
E = k+d+e = 59
F = l+f+g = 83
G = l+g+h = 79
H = l+h+i = 61
I = l+i+j = 67
J = l+j+f = 97
K = a+f+g = 73
L = b+g+h = 71
M = c+h+i = 79
N = d+i+j = 73
O = e+j+f = 101
P = f+e+a = 61
Q = g+a+b = 37
R = h+b+c = 73
S = i+c+d = 61
T = j+d+e = 79
|
Same as with consecutive primes |
1.2) Consecutive primes
| Cube |
Dodecahedron |
Icosahedron |
Sum = 90
Faces: a:7 b:11 c:23 d:17 e:13 f:19
Vertices:
A = e+a+b = 31
B = e+b+c = 47
C = e+c+d = 53
D = e+d+a = 37
E = f+a+b = 37
F = f+b+c = 53
G = f+c+d = 59
H = f+d+a = 43
|
Sum = 276
Faces: a:5 b:7 c:11 d:23 e:37 f:17 g:31 h:29 i:13 j:43 k:19 l:41
Vertices:
A = k+e+a = 61
B = k+a+b = 31
C = k+b+c = 37
D = k+c+d = 53
E = k+d+e = 79
F = l+f+g = 89
G = l+g+h = 101
H = l+h+i = 83
I = l+i+j = 97
J = l+j+f = 101
K = a+f+g = 53
L = b+g+h = 67
M = c+h+i = 53
N = d+i+j = 79
O = e+j+f = 97
P = f+e+a = 59
Q = g+a+b = 43
R = h+b+c = 47
S = i+c+d = 47
T = j+d+e = 103
|
Sum = 710
Faces: a:3 b:5 c:7 d:11 e:13 f:17 g:19 h:23 i:29 j:31 k:43 l:73 m:47
n:41
o:67 p:53 q:61 r:71 s:37 t:59
Vertices:
A = p+b+c+d+q = 137
B = q+d+e+f+r = 173
C = r+f+g+h+s = 167
D = s+h+i+j+t = 179
E = t+j+a+b+p = 151
F = k+a+b+c+l = 131
G = l+c+d+e+m = 151
H = m+e+f+g+n = 137
I = n+g+h+i+o = 179
J = o+i+j+a+k = 173
K = k+l+m+n+o = 271
L = p+q+r+s+t = 281
|
Regarding the method he used, he wrote:
My program assigns a number of consecutive primes to a list.
In the consecutive case, the number is the number of faces, in the general
case, I extend the list with a selectable number of such primes. I then
then permute all primes but the first and lowest and pick the "Faces"
first primes.
Using a unique table for each polyhedra, I then compute the vertex sums
and check for primality.
If all vertices are prime, I output the solution.
As the body is fully symmetric, there is no idea to include the first
prime in the permutations. Either there is a solution including the lowest
prime, which will be found or there is none.
When all possible permutations are exhausted, I start with next prime and
repeat the procedure. This way solutions starting with higher lowest prime
will be found.
Finding a solution with distinct primes in corners is new twist.
It can be done, but will take a long time for the two larger polyhedra, as
a lot of solutions have to be tested.
Ray Opao wrote:
Q1.
CUBE--distinct primes
a = 23, b = 17, c = 11, d = 5, e = 13, f = 3, S = a+b+c+d+e+f = 72
CUBE--consecutive primes
a = 17, b = 11, c = 13, d = 7, e = 23, f = 19, S = a+b+c+d+e+f = 90
Q2.
Let d = 3, and a,b,c be primes greater than 3.
Then a = 1 or 2 mod 3; b = 1 or 2 mod 3; and c = 1 or 2 mod 3.
Then A = a+b+c can be shown in 8 possible ways:
(1) 1 + 1 + 1 = 0 mod 3
(2) 1 + 1 + 2 = 1 mod 3
(3) 1 + 2 + 1 = 1 mod 3
(4) 1 + 2 + 2 = 2 mod 3
(5) 2 + 1 + 1 = 1 mod 3
(6) 2 + 1 + 2 = 2 mod 3
(7) 2 + 2 + 1 = 2 mod 3
(8) 2 + 2 + 2 = 0 mod 3
If A is prime, then cases (1) and (8) are eliminated.
If B = a+b+d = a+b+3 is prime, then B = 1 or 2 mod 3.
Or a+b = 1 or 2 mod 3.
But as a+b = 0 mod 3 for cases (3), (4), (5), and (6), then the said cases
are eliminated, leaving cases (2) and (7).
If C = a+c+d = a+c+3 is prime, then C = 1 or 2 mod 3.
Or a+c = 1 or 2 mod 3.
But as a+c = 0 mod 3 for the remaining cases (2) and (7), then both cases
are also eliminated.
Hence, it is impossible for A to be prime if B and C are both prime.
Which is why no solution exists when d = 3.
