A cube of size 4 × 4 × 4 is divided into 16 equal squares per face, with numbers from 1 to 96 randomly assigned to these squares. An operation consists of taking two squares that share a vertex, summing their numbers, and rewriting this sum in one of the squares while leaving the other blank. After performing several such operations, only one number remains. Prove that regardless of the order of operations, the final remaining number is always the same. Also find this number.
I don't really understand the process involved. What does it mean for two squares to share a vertex? Are we talking about the 8 vertices of the cube? If so, we will not arrive at a single number, so the author must have something else in mind.
But no matter. If we do arrive at a single number via repeated additions, then surely it must be 1+2+...+64 = 64*65/2 = 2080. How can it be anything else? And obviously, the total of all 64 numbers is independent of the order of operations, because addition is commutative.
What's the problem? (spoiler)