You have an empty box which is a 11cm cube. You also have a supply of blocks. You have 54 2cm cubes, 24 3cm cubes and two 5cm cubes. Pack all 80 cubes into the box.

Both solutions presented so far (Dan Blume's and mine) do follow a pattern with six 6x6x5 sub-blocks in each of which there is a layer of 4 3-cm cubes and a layer of 9 2-cm cubes. A question then exists: are there any solutions that do not follow this pattern?

If any solution must actually follow this pattern, then how many variants are there?

The answer to that depends on whether you are considering a particular orientation for the box, or consider a solution to be the same even if it is rotated.

When orientation is considered, the first 5x5x5 cube can be placed in any of the four bottom corners of the box. The 6x6 area occupied by the lying-flat 6x6x5 sub-assembly must be adjacent to this (otherwise it would create a 1-cm-thick layer eventually), but either clockwise or counterclockwise from the 5x5x5 cube. From there, the pieces will fit in only one way. That's 4x2 choices so far.

Within each 6x6x5 subassembly, the 3-cm cubes can be either toward the outside (including bottom or top) of the box or toward the inside (leaving the 2-cm cubes to face the outside). As there are 6 of these subassemblies, that gives each of the 8 original choices 2^6 possibilities, so the number of solutions is 8*2^6 = 2^9 = 512 ways of packing the box.

However if the orientation of the larger box is not to count, there would be fewer distinct packings. The diagonal formed by the two 5x5x5 cubes can be placed in a standard orientation. When considering the placement of the 6x6 flat on the surface, these can be either clockwise or counterclockwise at a given vertex with the 5x5x5 cube, but in any packing, one will be clockwise and the other counterclockwise, so there is no clockwise version vs. counterclockwise version.

The orientations of 3-cm out or 2-cm out for the 6x6 explosed flats, about the clockwise vertex can be (said regarding the 3-cm side):
1) All outward
2) all inward
3) Two outward
4) two inward.

In the case of "all outward" or "all inward", there are the same four possibilities at the other end of the diagonal axis, accounting for a total of 8 packings so far.

In the case of "two outward" or "two inward" the "all outward" or "all inward" at the other end constitute one way each, accounting for 4 more packings. But the "two outward" or "two inward" orientations at the other end also, could themselves be oriented 3 ways each with regard to where the odd one is in relation to the odd one at the other end, so 2x2x3=12 is added to the total. Thus the final total is 8+4+12 = 24 ways of packing the cubes, when orientation of the whole box is not considered.

The only remaining consideration is whether there are solutions which do not have the 6 identical 6x6x5 subassemblies.

Posted by Charlie on 2004-01-05 09:26:12