We are assembling 5 identical unit cubes in such a way that 4 of them form a stack 4 units high, and the fifth one connects to one face of one of the inner cubes. Can a solid rectangular prism be constructed with fewer than 20 of these shapes?
If such a rectangular parallelepiped exists, its volume is divisible by 5, since it is composed entirely of elements with a volume of 5. Therefore, at least one edge of the parallelepiped must be 5 units or a multiple of 5.

Let's try the simplest possible shape: let the parallelepiped be one layer thick, with a width of 5, and for now, let's not specify its length. Start filling it with our elements as shown in Figure 1.

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Figure 1

Notice that the boundary line of the filled part is symmetrical with respect to the marked point. Therefore, we can "push together" the filled part and its mirror image (Figure 1), forming a compact $1 \times 5 \times 10$ parallelepiped (Figure 2).

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Figure 2

For the assembly, we used a total of 10 elements, so the answer to the question is: yes, a solid parallelepiped can be assembled from fewer than 20 such elements.

Another type of assembly is shown in Figures 3 and 4, where a $3 \times 4 \times 5$ parallelepiped was made from 12 elements, and a $4 \times 4 \times 5$ parallelepiped from 16 elements.

$1985-11-387-1$. eps Figure 3 $1985-11-387-2$. eps Figure 4

Note. In our article "Puzzle Games" (p. 341), you can read about further interesting aspects related to this problem. Since a solid $5 \times 5 \times 5$ cube can also be assembled from 25 instances of the element, it is certainly not true that the following "obvious" statement, which several of our solvers also claimed, is correct: "If a solid, symmetrical parallelepiped can be assembled from non-symmetrical elements, then the elements must be placed symmetrically."