The real numbers $p, q, r$ are pairwise distinct, and they satisfy

$$
q=p(4-p), \quad r=q(4-q), \quad p=r(4-r) .
$$

What can be the value of $p+q+r$?
Suppose that

$$
q=p(4-p), \quad r=q(4-q), \quad p=r(4-r)
$$

Let $x=q-2, y=p-2, z=r-2$. Then (1) becomes:

$$
x=2-y^{2}, \quad y=2-z^{2}, \quad z=2-x^{2}
$$

$p, q, r$ are distinct if and only if $x, y, z$ are distinct. Suppose $x=y$. Then $x=2-x^{2}$, which gives $x=1$ or $x=-2$. If $x=1$, then $x=y=z=1$; if $x=-2$, then $x=y=z=-2$. We get the same result if $x=z$ or $y=z$. Therefore, $x, y, z$ are distinct if and only if $x \neq 1$ and $x \neq -2$.

From (2), we have:

$$
x=2-y^{2}=2-\left(2-z^{2}\right)^{2}=2-\left(2-\left(2-x^{2}\right)^{2}\right)^{2}
$$

Expanding the parentheses, we get:

$$
x^{8}-8 x^{6}+20 x^{4}-16 x^{2}+x+2=0
$$

It is easy to see that the above equation has roots $x=1$ and $x=-2$. Since according to the previous, neither $x-1 \neq 0$ nor $x+2 \neq 0$, we can divide (3) by $(x-1)(x+2)$; we get:

$$
x^{6}-x^{5}-5 x^{4}+3 x^{3}+7 x^{2}-x-1=0
$$

Factoring this, we get:

$$
\left(x^{3}-3 x-1\right)\left(x^{3}-x^{2}-2 x+1\right)=0
$$

Thus, either $x^{3}-3 x-1=0$ or $x^{3}-x^{2}-2 x+1=0$. Since our transformations are equivalent, if $x$ is a root of one of the above two equations, and $y=2-x^{2}, z=2-y^{2}$, then (2) holds for $x, y, z$.

Now we can determine the value of $p+q+r$.

$$
p+q+r=x+y+z+6, x+y+z=x+\left(2-x^{2}\right)+\left(2-\left(2-x^{2}\right)^{2}\right)=-x^{4}+3 x^{2}+x
$$

Case I: If $x^{3}-3 x-1=0$, then

$$
x+y+z=-x^{4}+3 x^{2}+x=-x\left(x^{3}-3 x-1\right)=0, p+q+r=x+y+z+6=6
$$

Case II: If $x^{3}-x^{2}-2 x+1=0$, then

$$
x+y+z=-x^{4}+3 x^{2}+x=-(x+1)\left(x^{3}-x^{2}-2 x+1\right)+1=1, p+q+r=x+y+z+6=7
$$

Therefore, the value of $p+q+r$ can be 6 or 7.

Remark. Many solved for the possible values of $q$ using the formula for solving cubic equations, and from there calculated the value of $p+q+r$.