Saudi Arabian Mathematical Competitions

1) For $k=5$, we need to prove that for all $m, n$ satisfying $5 n + m \mid 5 m + n$ then $n \mid m$. Note that $5 n + m \leq 5 m + n$ or $n \leq m$, then $1 \leq \frac{5 m + n}{5 n + m} < 5$.

Then $A = \frac{5 m + n}{5 n + m} \in \{1, 2, 3, 4\}$. We consider some cases:

- If $A = 1$ then $m = n$.
- If $A = 2$ then $5 m + n = 10 n + 2 m \Leftrightarrow m = 3 n$.
- If $A = 3$ then $5 m + n = 15 n + 3 m \Leftrightarrow m = 7 n$.
- If $A = 4$ then $5 m + n = 20 n + 4 m \Leftrightarrow m = 19 n$.

So in all cases, we always have $n \mid m$, which implies that $k = 5$ is a nice number.

2) We can directly check that $k = 2$ is a nice number. Consider some nice number $k > 2$. By a similar way, we can check that $n \leq m$ and
$$1 \leq \frac{k m + n}{k n + m} < k.$$
Thus $A = \frac{k m + n}{k n + m} \in \{1, 2, 3, \ldots, k-1\}$. In case $A = 2$, we have
$$\frac{k m + n}{k n + m} = 2 \Leftrightarrow k m + n = 2 m + 2 k n \Leftrightarrow \frac{m}{n} = \frac{2k - 1}{k - 2}.$$
We must have $\frac{m}{n} \in \mathbb{Z}^+$ then $\frac{2k - 1}{k - 2} = 2 + \frac{3}{k - 2} \in \mathbb{Z}^+$. Since $k > 1$, this means that $k - 2 \in \{1, 3\}$ or $k \in \{3, 5\}$.

It is easy to check that $k = 3$ is also a nice number. Therefore, all nice numbers are $2, 3, 5$.