The puzzle is as follows:

The figure from below belongs to a didactical toy which is comprised of 32 congruent wood pieces as indicated in the figure. Each piece is made up by three cubes whose edges measure 3 centimeters. If by using only these you want to make the maximum amount of compact cubes as possible. How many of these pieces you will not use?

A picture, labeled "Fig. 1. Sketch of the problem", showing three cubes in an L shape.

The choices given are as follows:

  1. 5 pieces
  2. 3 pieces
  3. 4 pieces
  4. 9 pieces

This puzzle it seems to stem from an old APA IQ exam from mid 1990s on psychometry for intelligence which is based on Leon Thurstone's and Raymond Catell IQ test timed cards.

I'm not sure how to solve it.

The question does not ask directly for the number of pieces, but instead in reverse for the number of pieces which will be unused or discarded to make cubes with only the pieces shown.

This part requires spatial visualization, which I lack, so an auxiliary drawing or figure would be helpful for any solutions.

In other words how to make a cube by using only the piece shown? The source doesn't indicate that the cubes can be split and rearranged in a line or any other shape. So I'm assuming that the intended solution does not modify the shape given.

Moving just this piece around I made a 3×3 cube. That is using six of these pieces.

In total it would be 27 cubes.

There would be 32-27=5 unused cubes. But this begs a question. Can these cubes be split or not?

Had this toy be composed of single cubes, I wouldn't bothered to assemble them in any way. I don't know how to correctly interpret the problem

Since 5 is one of the choices (#1) given, it might be the answer. But I can't say for sure. Perhaps another way to see this problem exists.

Therefore I'm requesting help. Please include drawings in answers so I can better visualize the solution to this puzzle.


You already gave the solution.

Regardless of how you build the cubes, they must consist of a multiple of 3 unit cubes. That allows only size 3 with 27 unit cubes or size 6 with 216 unit cubes.
Clearly the large option requires too many pieces so only cubes with 27 unit cubes or 9 pieces can be made.
So if a solution exists, you use 3x9 pieces and you have 5 pieces left.

Now how can you make a size 3 cube?

First join 3 pieces side by side to make a "sofa". Then assemble 3 blocks of 1x2x3 unit cubes using 2 pieces each. Put one block behind the sofa, one block on top, and one block sitting across the sofa.

enter image description here

  • $\begingroup$ I'm sorry for the late reply. But I'm still confused. If $9$ pieces are used to make the cube wouldn't this means that $32-9=23$ are the unused pieces?. The latter makes much sense than what just five. Isn't it?. Can you please explain this part. $\endgroup$ – Chris Steinbeck Bell Mar 21 at 2:23
  • $\begingroup$ You can make 3 cubes. Each cube uses 9 pieces. 32 - 3x9 = 32 - 27 = 5. $\endgroup$ – Florian F Mar 22 at 7:43

A cube of side length $n$ consists of $n^3$ cubes, and for such a cube to be buildable from these pieces, the volume $n^3$ of the cube needs to be evenly divisible by the volume $3$ of the pieces, therefore also $n$ needs to be divisible by $3$.

If the side length of an assembled cube is not $3$, it is therefore at least $6$, so it has volume at least $216 = 3 \cdot 72$ so it would need at least $72$ pieces to build. But we don't have that many pieces, so all we can build is cubes of side length $3$, requiring $9$ pieces each. After $3$ such cubes we have used $27$ pieces and are left with $5$ remaining pieces.


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