19. Consider a square puzzle, with $n$ pieces per side. Each piece has 4 sides, and on each side there can be either a hole or a protrusion, or it can be flat (when it is on the edge). Which of the following statements is false?

(A) If the total number of pieces is a multiple of 21, then the number of holes must also be a multiple of 21.

(B) The total number of holes is equal to the number of protrusions.

(C) If the total number of pieces is a multiple of 25, then the number of holes must also be a multiple of 25.

(D) The total number of protrusions is a multiple of 4.

(E) If the total number of pieces is odd, then the number of pieces that do not have flat sides must also be odd.
(19) The correct answer is $(C)$.

First, let's observe that $(B)$ is clearly true: the number of holes and the number of protrusions are equal (since for each hole there is one and only one protrusion) and their number coincides with the number of internal sides in the completed puzzle, identifying as a single side the sides of two pieces that fit together. This number, indicating with $n$ the number of pieces arranged along each side, is equal to $N=\frac{4 n^{2}-4 n}{2}=2 n(n-1)$.

Now let's examine the other statements.

$(A)$ is true: the total number of pieces $n^{2}$ is a multiple of 21 if and only if $n$ contains the factors 3 and 7; in such a case, $N$, which is a multiple of $n$, will also be a multiple of 21.

$(D)$ is also true: indeed, in the number $N=2 n(n-1)$, one of the two factors $n$ and $n-1$ must be even, being consecutive integers, so $N$ is certainly a multiple of 4.

Similarly, $(E)$ is true: the pieces without flat sides are those that do not face the edge of the puzzle, and their number is $(n-2)^{2}$; if the total number of pieces $n^{2}$ is odd, then $n$ is also odd, and so are $n-2$ and $(n-2)^{2}$.

Finally, let's see that $(C)$ is false. For example, take $n=5$: in this case, the number of pieces is 25, while $N=40$.

Question proposed by Andrea Barbarino.