Romanian Mathematical Olympiad

Notice that $A_n\subset \mathbb{Z}$ for all $n\in \mathbb{N}$, $n\ge 2$.

a) The elements of $A_2$ satisfy the inequalities $x-2<2x\le x$. By inspection, we obtain $A_2=\{-1,0\}$. The elements of $A_3$ satisfy the inequalities $5x-12<6x\le 5x$. By inspection, we have $A_3=\{-7,-5,-4,-3,-2,0\}$, so $A_2\cup A_3=\{-7,-5,-4,-3,-2,-1,0\}$.

b) For $n\ge 4$ and $x\in A_n$ we have $x\le \left(\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}\right)x$. From $\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}>1$ we obtain $x\ge 0$. For $x,n\in \mathbb{Z}$ and $n\ge 2$ we get $\lfloor \frac{x}{n}\rfloor \ge \frac{x-(n-1)}{n}$. Therefore, if $n\ge 4$ and $x\in A_n$, then $$x\ge \lfloor \frac{x}{2}\rfloor +\lfloor \frac{x}{3}\rfloor +\lfloor \frac{x}{4}\rfloor \ge \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4},$$ implying $x\le 23$. Hence the set $A$ is upper bounded. Because $A\subset \{-5,-4,\dots ,23\}$ and $23\in A_4$, then $\max A=23$.