Problems & Puzzles: Puzzles

Puzzle 230. Primes and a tower of cubes

For those that found a kind of difficult the previous puzzle (229) maybe will find this new puzzle more affordable.

Here you are invited to allocate the numbers from 1 to N, in the  N nodes of a tower of C cubes,  such that the sum of the numbers of the two nodes at the extreme of each edge (horizontal and vertical) is a prime number.

Evidently in a tower like this, N = 8 + 4(C-1) and the number of edges E is E = 12 + 8(C-1). E is also the number of prime-sums formed in the process of allocation.

Example. For C= 10, N = 44, E = 84 (*)

 5 12 25 42 38 41 18 29 35 32 11 8 44 39 20 23 27 40 33 14 34 19 28 9 37 24 43 10 36 17 30 31 7 6 13 16 22 1 4 15 21 2 3 26

Questions:

1) Find the smallest tower (the smallest C) that has solution.
2) Find a solution for C=20, 50, 100.

_______
(*) A tower of C cubes can be described as a matrix of (C+1) rows and 4 columns. In this model every element of the matrix represents a node, and the four elements of a row represent  the four nodes of an horizontal face of a cube.

Solution: Giovanni Resta solved this puzzle completely:

"...smallest C is 5: (which is also a torus, that is last and fist face can be connected (10+21,16+13,15+4,22+7). I have also counted the number of solutions in this case, but I have the data at home (so next week, since I'm leaving for half-a-week).

21 16 15 22
20 3 14 9
11 8 5 2
6 23 24 17
1 18 19 12
10 13 4 7

He also sent the solutions for C=20, 50 & 100 (too large to be exposed explicitly here)

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