(In reply to re: an answer
by Ady TZIDON)
There seems to be a flaw in my previous proposed solution. Though using it as a basis, the number of arrangements of the cube to indicate the bottom unseen number may be 16.
The cube can be arranged to indicate up to 6 sequential values less than, between, or greater than a given the five visible numbers. Yet, these sequential strings of values need overlap to allow for certainty of designation, thus the number of possibilities is 6+5 + 5, for the visible numbers = 16. Both Bob and Alex would need to understand and follow the same scheme to implement this, which I will will agree with Ady that it would be difficult. Yet, I believe is possible.
Is there a solution that can account for a larger value of N?
Edited on July 16, 2013, 11:59 pm
Posted by Dej Mar
on 2013-07-13 05:42:25