\section*{Problem 1 - 211031}

Determine all triples \((a, b, c)\) of natural numbers with the following properties!

(1) It holds that \(0 < a \leq b \leq c\).

(2) In a cuboid with length \(a \mathrm{~cm}\), width \(b \mathrm{~cm}\), and height \(c \mathrm{~cm}\), the sum of all edge lengths is equal to the volume in cubic centimeters.
}

(2) means that the equation \(4(a+b+c)=a b c\) needs to be solved. Due to (1), \(a b c=4(a+b+c) \leq 12 c\), thus \(a b \leq 12\) and therefore \(a \leq 3\).

Case 1: \(a=1\). Then the equation \(b c=4(1+b+c)\) with \(b \leq 12\) needs to be solved.

Case 1.1: \(b=4\). Then we get \(4 c=4(5+c)\), which has no solution.

Case 1.2: \(b \neq 4\). It holds that \(c=\frac{4(b+1)}{b-4}\). Since \(g g T(b+1, b-4)=g g T(b-4,5) \in\{1,5\}\), \(c\) can only be a natural number if \(b-4\) is a divisor of \(4 \cdot 5=20\). Indeed, the remaining possibilities \(8=12-4 \geq b-4=1,2,4,5\) yield the solutions \((1,5,24),(1,6,14),(1,8,9)\) and \((1,9,8)\), where the last one is discarded because \(c<b\).

Case 2: \(a=2\). Then the equation \(2 b c=4(2+b+c)\) or \(b c=2(2+b+c)\) with \(2 \leq b \leq 6\) needs to be solved.

Case 2.1: \(b=2\). Then \(2 c=2(4+c)\), which has no solution.

Case 2.2: \(b \neq 2\). We get \(c=\frac{2(b+2)}{b-2}=2+\frac{8}{b-2}\), which only yields natural numbers if \(b-2 \mid 8\). Given \(4=6-2 \geq b-2\), we obtain the solutions \((2,3,10),(2,4,6)\) and \((2,6,4)\), where the last one is again discarded because \(c<b\).

Case 3: \(a=3\). Then the equation \(3 b c=4(3+b+c)\) with \(3 \leq b \leq 4\) needs to be solved.

Case 3.1: \(b=3\). Then \(9 c=4(6+c)\) or \(5 c=24\), which has no solution in natural numbers.

Case 3.2: \(b=4\). Then \(12 c=4(7+c)\) or \(3 c=7+c\), thus \(c=\frac{7}{2} \notin \mathbb{N}\), so there is no solution here either.

In total, we obtain the following five solution triples:

\[
L=\{(1,5,24),(1,6,14),(1,8,9),(2,3,10),(2,4,6)\}
\]