## Task 6 - 120936

a) Determine the number of all different triples $(k, n, m)$ of natural numbers $k, n, m$ for which $k \cdot n^{2} \cdot(2 m+1)=3808$ holds.

b) List all the triples mentioned in a) for which the product $knm$ takes the smallest value.
a) It holds that $3808=2^{5} \cdot 7 \cdot 17$. Since $2 m+1$ must be an odd divisor of 3808, it can only be 7, 17, or $119=7 \cdot 17$. Therefore, $m \in\{3,8,59\}$.

For $n^{2}$, the possibilities are $n^{2}=4^{2}=2^{4}$, $n^{2}=2^{2}$, or $n^{2}=1^{2}$ (independent of how $m$ is chosen). Each of these choices uniquely determines $k$, so there are $3 \cdot 3=9$ possibilities.

b) If $n=1$ or $n=2$ is chosen, then $k$ contains the missing factor of 4 squared. To obtain the smallest possible product, $n=4$ must be chosen. Among the remaining possibilities $(k, m, n) \in\{(34,4,3),(14,4,8),(2,4,59)\}$, the product $34 \cdot 4 \cdot 3=408$ has the smallest value.