Consider the cuboid of dimensions 19 x 13 x 7 whose volume is 1729, the Hardy-Ramanujan number. What is the least number of smaller cuboids into which it can be dissected so that the resulting pieces can be reassembled, first as a unit cube and a cube of side 12, and next as a cube of side 10 and another of side 9?
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$\begingroup$ I have 9 cuboid-shaped pieces for the Ramanujan cuboid and the 9 and 10 cubes. I might be able to break the 10x10x7 piece into few enough pieces to get a fewer than 20 piece decomposition as desired. Gerhard "Using Divide And Conquer Approach" Paseman, 2016.06.03. $\endgroup$– Gerhard PasemanJun 3, 2016 at 17:15
2 Answers
J. H. Cadwell, A Three-Way Dissection Based on Ramanujan's Number, The Mathematical Gazette Vol. 54, No. 390 (Dec., 1970), pp. 385-387, DOI: 10.2307/3613865 gives a 12-piece dissection.
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$\begingroup$ Indeed this is a find and thank you for it. As the original post mentioned dissection into cuboids (which I assume are rectangular prisms), do you have a reference for this kind of dissection also? Gerhard "Likes Straight Edge Saw Cuts" Paseman, 2016.06.03. $\endgroup$ Jun 3, 2016 at 23:42
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$\begingroup$ @Gerhard, sorry, no (so I guess my answer doesn't actually answer the question as posed). $\endgroup$ Jun 4, 2016 at 1:33
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$\begingroup$ A very useful reference indeed (thanks @GerryMyerson), but I would still like to know the least number of cuboids (rectangular prisms) required. $\endgroup$ Jun 4, 2016 at 1:35
This is a partial answer, as I have not fully explored assembly into a cube of side 12. It is a decomposition into 9 cuboids to get the Ramanujan cuboid and the cubes of sides 9 and 10. Maybe someone can break the large piece into 10 or fewer cuboids to finish the answer and give an upper bound.
As in the paper in Gerry Myerson's answer, I start with two large pieces of the cuboid, of sizes 9x7x13 and 10x7x13, and break a large piece off each of these (9x9x7 and 10x10x7), leaving 9x7x4 and 10x7x3. The latter piece I leave alone, and the former I divide in half leaving a 9x7x2 piece from which I need to make a 3x3x10 cuboid and a 9x2x2 cuboid. The latter can be cut as a single piece from the 9x7x2 cuboid, giving the third piece for the 9-cube. The remaining 9x5x2 block can be split into two 3x5x2 and two 1x5x3 cuboids to complete the 6 piece dissection of the 10-cube.
All the small pieces from the 10-cube dissection fit in the shell formed by removing a corner 9-cube from a 12-cube. If we need to cut the 10x10x7 cuboid to get the remaining pieces, we will need at least five pieces: four to extract the 1-cube, and as one of those pieces is too thick to share room with a 9x9x7 cuboid in a 12-cube, it will need to be cut. Seeing a published dissection having 12 non cuboid pieces, I will be surprised if any cuboidal dissection exists with 14 pieces. I expect a dissection of the above (all the listed pieces above, except breaking the 10x10x7 into smaller cuboids) may be achieved with 18 pieces, but I haven't gotten there yet.
Gerhard "Is Feeling Somewhat Cubist Today" Paseman, 2X16.-6.X4
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$\begingroup$ Indeed, I think breaking the 10x10x7 into two 10x9x3 cuboids, three 9x2x1 cuboids, two 9x3x1 cuboids, a 10x3x1 cuboid, three 2x3x1 cuboids, one 1x1x3 cuboid, and 1-cube gives a 21 piece dissection into cuboids which can probably be improved. Gerhard "Leaves The Improving To Others" Paseman, 2016.06.04. $\endgroup$ Jun 4, 2016 at 17:53