John has 77 boxes each having dimensions 3x3x1. Is it possible for John to build one big box with dimensions 7x9x11?
I'm leaning towards no, but I would like others opinion.
The answer is no; John can't even fill up the topmost $7\times 11\times 1$ slice of the $7\times 11\times 9$ box. Consider just the top $7\times 11$ face of this box; look just at this face and ignore the rest of the box. A solution to the problem would fill up this $7\times 11$ rectangle with large $3\times3$ rectangles and small $3\times 1$ rectangles. But $7\times 11$ is not a multiple of $3$.
The volume of the big box is $V_B = 7\cdot 9 \cdot 11 = 693$, the total volume of the small boxes is $V_b = 77 \cdot 3 \cdot 3 \cdot 1 = 693$.
This means the volume of the small boxes is sufficient and we need to use all small boxes.
Let us try to model this problem Tetris style:
This is a search space of $77\times 7 \times 9 \times 3 = 14553$ drop configurations. Not that much for a machine.
We could avoid the drop simulation and instead have $c_z$ as another choice. This would enlarge the search space to $77\times 7 \times 9 \times 11 \times 3 = 160083$ configurations. In both cases we need to check that boxes do not intersect.
This should be sufficient to code a solver which visits all configurations of the search space (brute force) and will answer the question by either listing feasible configurations or reporting that there is no solution.
Note: I submitted this before MJD published a counter argument.