Dean Hoffman (1978).
27 bricks of A×B×C into a cube with sides of A+B+C.
Then, (A+B+C)/4 < A < B < C.
Dean Hoffman (1978).
27 bricks of A×B×C into a cube with sides of A+B+C.
Then, (A+B+C)/4 < A < B < C.
| Pieces | 27 | All pieces are congruent. |
| Size | 4×5×6 | |
| Goal | 15×15×15 | |
| Holes | 135 | |
| Solutions | 21 |
Donald Knuth (2004).
28 bricks of A×B×C into a cube with sides of A+B+C.
Then, (A+B+C)/4 = A < B < C.
This is NOT Hoffman's Packing Puzzle.
| Pieces | 28 | All pieces are congruent. |
| Size | 3×4×5 | |
| Goal | 12×12×12 | |
| Holes | 48 | |
| Solutions | 20 |