Five, there is a rectangular prism with length, width, and height being positive integers $m, n, r (m \leqslant n \leqslant r)$, respectively. After the surface is painted red, it is cut into cubes with an edge length of 1. It is known that the sum of the number of cubes without red and the number of cubes with two red faces, minus the number of cubes with one red face, equals 1985. Find the values of $m, n, r$.
Let $m>2$, according to the problem, the number of cubes without red is
$$
k_{0}=(m-2)(n-2)(r-2) ;
$$

The number of cubes with one red face is
$$
k_{1}=2(m-2)(n-2)+2(m-2)(r-2)+2(n-2)(r-2) \text {; }
$$

The number of cubes with two red faces is
$$
k_{2}=4(m-2)+4(n-2)+4(r-2) ;
$$

Thus, we have
$$
k_{0}-k_{1}+k_{2}=1985 \text {. }
$$

That is, $(m-2)(n-2)(r-2)+4[(m-2)+(n-2)+(r-2)]$
$$
-2[(m-2)(n-2)+(m-2)(r-2)+(n-2)(r-2)]=
$$
1985.

Or $[(m-2)-2][(n-2)-2][(r-2)-2]$
$$
\begin{array}{l}
=1985-8=1977 \text {. } \\
(m-4)(n-4)(r-4)=1977 \text {. } \\
\because \quad 1977=1 \times 3 \times 659=1 \times 1 \times 1977 \\
=(-1) \times(-1) \times 1977 \text {, } \\
\therefore \quad m-4=1, n-4=3, r-4=659 \text {. } \\
\end{array}
$$

Or $m-4=1, n-4=1, r-4=1977$.
Or $m-4=-1, n-4=-1, r-4=1977$.
Therefore, $m=5, n=7, r=663$.
Or $m=5, n=5, r=1981$.
Or $\quad m=3, n=3, r=1981$.
Let $m=1$, at this time $n=1$, no solution, $\therefore \quad n \geqslant 2$.
According to the problem, the number of cubes without red is $k_{0}=0$, the number of cubes with one red face is $k_{1}=0$,
The number of cubes with two red faces is
$$
k_{2}=(n-2)(r-2) .
$$

Thus, we have $k_{0}+k_{2}-k_{1}=k_{2}=1985$.
That is $(n-2)(r-2)=1985=5 \times 397$.
$$
\begin{array}{l}
\therefore \quad n-2=5, r-2=397, \\
\text { or } \quad n-2=1, r-2=1985 . \\
\therefore \quad m=1, n=7, r=399,
\end{array}
$$

Or $m=1, n=3, r=1987$.
Let $m=2$, then $k_{0}=0, k_{1}$ and $k_{2}$ are both even,
Clearly, $k_{0}+k_{2}-k_{1} \neq 1985$.
In summary, there are five sets of $m, n, r$ that meet the problem's requirements:
$$
\begin{array}{l}
m_{1}=5, n_{1}=7, r_{1}=663 ; \\
m_{2}=5, n_{2}=5, r_{2}=1981 ; \\
m_{3}=3, n_{3}=3, r_{3}=1981 ; \\
m_{4}=1, n_{4}=7, r_{4}=399 ; \\
m_{5}=1, n_{5}=3, r_{5}=1987 .
\end{array}
$$