Cube Corner Puzzle (via MindYourDecisions on YouTube)

Let the numbers printed on the $XZ$ faces of the cube be $a, c$ , on the $YZ$ faces be $b, d$ and on the $XY$ faces be $e, f$ respectively. Then

$abf+adf+bcf+cdf+abe+ade+bce+cde=154\implies (a+c)(b+d)(e+f)=2\times 7\times 11$ .

Since $2,7,11$ are primes, we can assume the three factors to take these three values. Hence,

$a+b+c+d+e+f=2+7+11=\boxed {20}$ .

So we know that opposite faces cannot be part of the same expression in the term for a vertex, and since the question asks for positive integers, we are looking for a way to factorize the expression of the sum of the vertices. I'll group the opposite sides of the cube in pairs: (x, b), (y,a), (z,w) where x, y, z, w, a, b are integers. In order to include every single possibility of a term for a vertex and exclude all the possibilities for opposite terms to be grouped together we will group the opposite faces together, so: (x+b)(y+a)(z+w). We see that if we distribute this equation it gives us 8 terms for the 8 vertices, so that must be the factorized form. So (x+b)(y+a)(z+w) = 154 and 2 × 7 × 11 = 154, so 2 + 7 + 11 = 20. The answer is 20.