The puzzle is as follows:

Assume that you have this peculiar toy. This toy is composed by many plastic pieces which is shown in the figure from below, all of them are cubes. You can use as many as you please. The task is that with them you must make a figure similar to type 3. How many of these pieces at minimum you require to do so?. The condition is that you must use a piece of each type.

Sketch of the problem

The choices given are:

  1. 12 pieces
  2. 6 pieces
  3. 5 pieces
  4. 8 pieces

This puzzle comes from an old mid 1980s APA IQ test timed cards belonging to Leon Thurstone's/Raymond Catell's psychometric exam.

My spatial abilities are not good, and I don't know if there is a procedure for how to solve these without having the actual figure.

My best guess is that to make the type three figure you must have a tower which is in the ratio of 1 to 2.

I could use type 1 two times and put them one over the other making a compact box of 2×3×2 cubes.

Then I would put two type 2, one next to the other, and finally type 3 two times again.

Thus I'm getting a set of 6 pieces. But the resultant figure isn't exactly similar to type three. Can someone help me? It would be really helpful for answers to include a drawing as it would help me to better visualize the final figure.


1 Answer 1


You put two pieces of type 1 together to make a 2x3x2 brick.

You can then put one piece of type 2 on one end to make it a 2x4x2 brick. This is the shape we want, except that we haven't used any of the type 3 pieces. However, you can replace one of the type 1 pieces by three type 3 pieces.
So now you have a 2x4x2 block made of one type 1, one type 2, and three type 3 pieces.

So the answer is

5 pieces.

Can it be done with less?

The volume of the goal shape is 2x4x2=16 cubes. We need to use one of each piece, which is 6+4+2=12 cubes in volume. We used two extra type 3 pieces to make up the difference. The only way to get fewer pieces is if it were possible to use a single extra type 2 piece instead, and no extra type 3 pieces. This is fairly obviously impossible (the gap in the middle of the type 1 piece can only be filled using the type 3 piece, and then there is no way to finish the 2x4x2 block with two type 2 pieces).

  • $\begingroup$ Thanks for that. Although I requested a drawing to be included in the answer, your words are easy to understand on this one. $\endgroup$ Commented Mar 22, 2021 at 20:26

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