\section*{Problem 1 - 221031}

Determine all quintuples \((x, y, z, u, v)\) of natural numbers for which

\[
0 < x \leq y \leq z \leq u \leq v \quad \text{and} \quad x + y + z + u + v = x \cdot y \cdot z \cdot u \cdot v
\]

holds!
}

There apparently is no solution where \(x=y=z=u=v\). Therefore, \(5 v > x + y + z + u + v = x y z u v\), so \(x y z u < 5\). Four cases remain:

1. Case: \(x y z u = 1\).

This is only possible if \(x = y = z = u = 1\). Thus, \(1 + 1 + 1 + 1 + v = 1 \cdot 1 \cdot 1 \cdot 1 \cdot v\) must hold, which is impossible.

2. Case: \(x y z u = 2\).

Then \((x, y, z, u) = (1, 1, 1, 2)\) and hence \(1 + 1 + 1 + 2 + v = 2 v\), so \(v = 5\).

3. Case: \(x y z u = 3\).

Then \((x, y, z, u) = (1, 1, 1, 3)\) and hence \(1 + 1 + 1 + 3 + v = 3 v\), so \(v = 3\).

4. Case: \(x y z u = 4\).

Then \((x, y, z, u) = (1, 1, 2, 2)\) or \((x, y, z, u) = (1, 1, 1, 4)\).

In the first case, \(1 + 1 + 2 + 2 + v = 4 v\) implies \(v = 2\). In the other case, \(1 + 1 + 1 + 4 + v = 4 v\) contradicts \(v \in \mathbb{N}\).

In total, we obtain that \((1, 1, 1, 2, 5)\), \((1, 1, 1, 3, 3)\), and \((1, 1, 2, 2, 2)\) are all the desired quintuples.