SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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

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

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

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 $n\le m$ and $$1\le \frac{km+n}{kn+m}<k.$$ Thus $A=\frac{km+n}{kn+m}\in \{1,2,3,\dots ,k-1\}$. In case $A=2$, we have $$\frac{km+n}{kn+m}=2\iff km+n=2m+2kn\iff \frac{m}{n}=\frac{2k-1}{k-2}.$$ 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 $2,3,5$.