prime boxes of (61) Torsten Sillke, BI, complete: 20.01.93 1 2 1 1 2x5x 2p, 3p, ... {2..3}+2n 3x5x 2p, 4, 6, ... 2n 4x5x 2, 3, ... {2..3}+2n 5x5x 2, 4, 6, 8, 9p, ... {8..9}+2n 6x5x 2, 3, ... {2..3}+2n 7x5x 2, 4, 6, 7p, ... {6..7}+2n 8x5x 2, 3, ... {2..3}+2n 9x5x 2, 4, 5p, ... {4..5}+2n 3x10x 2, 4, 5, ... {4..5}+2n 3x15x 2, 4, 6, 8, 9p, ... {8..9}+2n 3x25x 2, 4, 6, 7p, ... {6..7}+2n 5x15x 2, 4, 5, ... {4..5}+2n 3x3x5x 2, 3p, ... {2..3}+2n Complete list of prime boxes (# = 5): 2x2x5, 2x3x5, 5x5x9, 5x7x7, 3x7x25. Complete list of prime hyperboxes (# = 1): 3x3x3x5. Impossible: 3x3xZ strip (two sides open) (A) 3x5xu (u odd) (B) 5x5x5, 5x5x7, 3x7x15 (C) Dist 2 Coloration: o . o . o one (61) matches at most one colored cube. . . . . . o . o . o (A) three is no 3x3xZ strip (two sides open): Take a 3x3x9 subblock from the strip. It contains 2x2x5 colored cubes. If you want to match these cubes in the strip you need at least 20 (61), which gives you a volume at least 100. On the other hand all Pentacubes are located inside a 3x3x11 subblock with volume 99. This is too small. (B) a 3x5xN strip (one side open) contains only 2x3x5 boxes: Color the 3x5 face with the Dist 2 Coloration. 6 cubes are colored. To match these cubes you need 6 (61). They have a volume of 30. All 6 (61) lay inside the first two layers. They have a volume of 30 too. So there is a split. (C) 5x5x5, 5x5x7, 3x7x15 : box : colored cubes, number of pentacubes 5x5x5 : 3x3x3 = 27 25 5x5x7 : 3x3x4 = 36 35 3x7x15: 2x4x8 = 64 63 the number of colored cubes is in each case too big. Annotations: The pentacube 61 has been considered by other puzzlers as well. Helmut Postl [2] wrote, that he found in the beginning of the 90th all prime packings of this pentacube by hand and found the coloration condition as well. Independently Fritz G"obel worked on this pentacube too [1]. He found the coloration condition and constructed the 5x5x9 box by hand. building blocks: 2 2 (61) build the figures: P: 2 2 Z: 2 2 2 2 2 2 2 3 3 | p p p p p p 3 3 3 3 | 2 p p 3 4 p p 5 6 6 7 7 3 3 3 3 | 2 2 3 3 2 2 3 3 6 6 7 7 3 3 | . . 3 3 . . . 3 7 . . . 7 7 . 3 3 3 3 3 | 8 8 3 9 9 8 6 3 9 8 6 6 7 7 8 3 3 3 3 3 | 8 5 5 9 9 8 5 2 2 8 6 6 2 8 8 3 3 3 3 | p 5 4 4 . p 5 2 1 . p p 2 1 . 3 3 3 3 | p p 4 4 . p p 4 1 . p p 1 1 . the 2x2x5 and 2x3x5 boxes: 2 P -> 2x2x5 2 P & Z -> 2x3x5 The 5x5x9 box: a a a x x b b c c is impossible but swap one 61 in the 'b' a a a x x b b c c block to the outside and you can pack 'x'. d d e e x b b c c d d e e x x f f f d d e e x x f f f 4 4 4 4' 3 3 q q q q The both primed numbers 4 9 9 3'9 3 3 q q q q can be switched which 9 9 8 q q gives the other solution. 0 . 8 8 8 8 p p p p 0 0 0 0 p p p p p p the 5x7x7 box: 6 (2x3x5) & 5 P & 1 1 -> 5x7x7 3 3 2 3 2 a a x x b b b The excess of 1 is used in the 2x2x5 subblock. a a x x b b b a a x x x c c d d d x x c c d d d e e f f g g g e e f f g g g e e f f the 3x7x25 box: a a a a a b b b b x x x x x x f f f f e e d d c c a a a a a b b b b x x x x x x f f f f e e d d c c c c d d e e b b f f x x x x f f f f f e e d d c c c c d d e e f f f f f x x x f f f f f e e d d c c c c d d e e f f f f f x x x x f f b b e e d d c c c c d d e e f f f f x x x x x x b b b b a a a a a c c d d e e f f f f x x x x x x b b b b a a a a a 9 9 8 8 7 7 9 9 8 8 7 7 6 6 5 5 4 4 9 3 3 a a 7 6 3 8 5 a 4 6 6 5 5 4 4 3 3 a a d d e e d d e e b b c b f e d f f b b c c g f c c g f 3 3 4 4 5 5 6 6 4 g 5 8 3 6 7 g g 3 3 9 4 4 5 5 6 6 7 7 8 8 9 9 7 7 8 8 9 9 the 3x3x3x5 hyperbox: L3x3x5 & 3 (2x3x5) -> 3x3x3x5 References: [1] Fritz G"obel, Prime Pentacube Packings, CFF 33, Feb. 1994, 24-25 (found the coloration condition, and the 5x5x9 box.) [2] Helmut Postl, letter from 23. July 1997 (found the coloration condition, and all prime boxes.)