TITLE: Dissecting Ramanujan´s Cuboid: 1729 = 19 x 13 x 7
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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