There is a non-negative integer written on all six faces of a cube. Integers on all possible three common adjacent faces (all faces are adjacent to each other) are multiplied and then they are added. The resulting sum is equal to 2022. Find the highest possible sum of all integers on the faces of the cube.
The cube has 3 pairs of opposite faces. If we pick 1 face from each pair, then these three faces meet at a vertex of the cube. In fact, the 8 ways to choose these 3 faces correspond to the 8 products.
Call the integers on the pairs of opposite faces $(a,b)$, $(c,d)$, and $(e,f)$. Using our earlier observation, the sum of all 8 products is $(a+b)(c+d)(e+f)$, where each product corresponds to choosing one element from each binomial.
Rewrite $x=a+b$, $y=c+d$, and $z=e+f$. Then $x$, $y$, and $z$ are nonnegative integers and $xyz=2022=2\cdot3\cdot337$. We seek to maximize $x+y+z$. Clearly this is achieved by $x=y=1$ and $z=2022$ for a total of 2024.
EDIT: This answers the question as it was during the first 20 seconds of its existence, and zeroes were not allowed. (In my (and Jaap's, see comments) opinion, this is the better form of the puzzle). For the answer to the question as it was amended to be, see @noedne's post.
To maximise the sum of the numbers making up a product, we'll want to minimise all numbers but one. (To create the edges of a rectangular prism from a strip of wood, you get maximum volume by creating a cube. Here we want to be as wasteful as possible: we want to use up maximum amount of wood for the given volume, so we want to build a thin, long shape instead)
So let's see if we can't have five sides with 1, and one with n
There are 8 corners in a cube, 4 of which are adjacent to n, and 4 that aren't. From this, we instantly see we are out of luck: 2022 isn't divisible by 4.
The same argument also rules out the "bigger numbers on two opposing edges, ones elsewhere" option.
So we have to multiply our big number by something. The minimum is then putting a 2 in the mix. It needs to be adjacent to the n, otherwise it's just a repeat of the previous attempt.
This time we get a corner product sum that's divisible by 6 (looking good)
$$2n + 2n + n + n + 2 + 2 + 1 + 1 = 6(n+1) = 2022$$
From this we can solve
$$ n=336 $$
which gives the maximum sum of all the faces:
$$336+2+1+1+1+1 = 342$$