# 3³+4³+5³=6³ Puzzle

A classic puzzle asks us to break a 6x6x6 cube into the smallest number of pieces which can be reassembled into 3 physically separate cubes of sizes 3, 4, & 5.

3³+4³+5³ =27+64+125 =216 =6³

An 8-piece solution is known to be minimal, because if any single fragment contains 2-or-more of the size-6 cube's corners, then it would be 6 units long in at least one direction, & therefore too large to fit inside any smaller cube. Thus each fragment must possess exactly one of the original cube's 8 corners.

One such solution uses an intact cube of size-3 as one of the 8 fragments, while 2 other fragments assemble into the size-4 cube, & the remaining 5 fragments build the size-5 cube.

A physically different solution allows the size-4 cube to remain intact, leaving 2 fragments to make the size-3 cube, & the remaining 5 fragments to build the size-5 cube.

Is there an 8-part solution that allows the size-5 cube to remain intact?
If not, then how do we establish an impossibility proof?

As a side note, what if we remove the requirement that the 3 smaller cubes must be physically separate? If we imagine stacking the size-4 cube above the size-5 cube, then placing the size-3 cube on top, then we have defined a single 12-unit-tall structure, containing the 3 smaller cubes, which are no longer physically separate from each other. An 8-piece solution is obviously possible, but can we build this from 7 fragments or less?

• I think you should separate the last paragraph into a new question. We prefer only one question per post so answers are clear what they apply to. Commented Jun 5 at 16:02
• Thanks for the puzzle and welcome to the site! Commented Jun 5 at 17:47
• Welcome to PSE (Puzzling Stack Exchange)! Commented Jun 5 at 18:05
• @Ross Millikan: Normally I'd agree, but based on previous experience, I believe the previous paragraphs are needed to provide sufficient clarity. Commented Jun 8 at 18:39
• Is it a requirement for this puzzle that all the pieces are polycubes? (The 8-piece minimum proof appears to assume this, and the solutions posted so far only use polycubes, but it is far from obvious to me that no simpler solutions exist that makes use of non-orthogonal cuts.)
– fljx
Commented Jun 11 at 7:41

Main question:

Secondary question:

• ... I'm impressed! I've been banging my head against this particular "brick wall" for more than 2 years! BRAVO! Commented Jun 9 at 0:28
• @Will Octagon Gibson: Eric Harshbarger has a site with a METRIC-TON of puzzles to explore! In his LEGO category, look for the entry "Cube Dissection" ... His 8pc. solution keeps the size-3 cube intact, & as a bonus, one of the 5 fragments that build the size-5 cube is itself a size-2 cube! ..... I've lost the link to the site containing the "intact size-4 cube" solution, but if you ask Col. George Sicherman, he should be able to help, as he pointed out the site to me originally. Commented Jun 9 at 1:27
• My own efforts were focused on treating the bottom 6 layers of this sculpture as a single massive fragment, & trying to minimize the parts needed from the remainder. Do you have any sense of whether your 7-fragment solution represents the minimum solution? Commented Jun 9 at 1:33
• @DMC_Run See updated images. Commented Jun 9 at 1:55
• @quarague I created and manipulated various blocks in a 3D CAD virtual environment, Tinkercad Commented Jun 10 at 11:03

An alternative 6-piece answer to the second question, which has the largest possible first and second piece:

First divide the 6-cube as follows. The white piece is a 5x5x5 cube plus a layer of 4x4, and all colored unit cubes are visible.

Then assemble the colored pieces into the 4x4x3 + 3x3x3 portion as follows. The colors of the piece(s) added in each image are green, blue, purple, and then red and yellow.