Baltic Way 2019

We show first that $k=32$ is such a number. Consider a partition $\{U,V\}$ of the set $\{2,3,\dots ,32\}$ where we may assume that $2\in U$. Towards contradiction, suppose that none of the parts contains numbers $a,b$ and $c$ with the desired property. As $2\in U$ and $2\cdot 2=4$, we have $4\in V$. Similarly, $4\cdot 4=16$ implies $16\in U$. Hence $2,16\in U$, but $2\cdot 8=16$, so $8\in V$. We have concluded that $2,16\in U$ and $4,8\in V$, but $2\cdot 16=32=4\cdot 8$, which implies that the number $32$ cannot be in any of the parts, which is a contradiction. Therefore, $k=32$ is a desired number.

We now prove that $k=32$ is actually the least number with the desired property. We form the partition $\{U,V\}$ of the set $K=\{2,3,\dots ,31\}$ in the following way. For any number $n\in {\mathbb{Z}}_{+}$, consider its prime factorization representation $n={\prod }_{i=0}^{k-1}{p}_i$, and put $\Omega (n)=k$. If $n<32=2^5$, then $n$ is necessarily a product of at most four primes (with repetitions counted) or $\Omega (n)\ge 4$. Put $$U=\{n\in K\mid \Omega (n)=1\text{ tai }\Omega (n)=4\}$$ and $$V=\{n\in K\mid \Omega (n)=2\text{ tai }\Omega (n)=3\}.$$ Let $a,b,c\in K$ be numbers with $ab=c$. We observe that $\Omega (c)=\Omega (a)+\Omega (b)$. On the other hand, we have $\Omega (c)\le 4$, as $c\in K$. If now $a,b\in U$, then necessarily $\Omega (a)=\Omega (b)=1$, which implies $\Omega (c)=2$ and $c\in V$. If instead of that $a,b\in V$, then $\Omega (a)=\Omega (b)=2$ and $\Omega (c)=4$, so $c\in U$. $\square$