At a vertex of a certain cuboid, the sum of the 3 edges meeting is $42 \mathrm{~cm}$, and its space diagonal is $27 \mathrm{~cm}$. What is the surface area?
First, we need to ensure that a rectangular parallelepiped meeting the requirements actually exists. If we denote the space diagonal by $d$ and the edges by $a, b, c$, then it is necessary that $a+b+c > d$. Furthermore, $a+b+c$ reaches its maximum value when $a=b=c$. (This simply follows from the well-known theorem that in a right-angled triangle with a constant hypotenuse, the sum of the legs is maximized when the legs are equal). In this case, $d = \sqrt{a^2 + a^2 + a^2} = a \sqrt{3}$, so $a = \frac{d}{\sqrt{3}}$ and thus $3a = d \sqrt{3}$. Therefore, the upper limit of $a+b+c$ is $d \sqrt{3}$.

In our problem,

$$
27 < 42 < 27 \sqrt{3} \approx 46.8
$$

and thus a rectangular parallelepiped meeting the requirements does exist.

The surface area of the existing rectangular parallelepiped is:

$$
F = 2ab + 2ac + 2bc
$$

According to the problem,

$$
a + b + c = 42
$$

and

$$
a^2 + b^2 + c^2 = 27^2
$$

Squaring (1),

$$
a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = 42^2
$$

Subtracting (2) from (3),

$$
2ab + 2ac + 2bc = 42^2 - 27^2
$$

Thus,

$$
F = 42^2 - 27^2 = 69 \cdot 15 = 1035 \text{ cm}^2
$$

Note: If 42 is mistakenly given as 47, then we would get $F = 47^2 - 27^2 = 74 \cdot 20 = 1480 \text{ cm}^2$, although in this case, the rectangular parallelepiped does not exist.

None of the solvers pointed out the existence of the rectangular parallelepiped. (This would have been an opportunity to earn a few valuable points in the competition.)