75th Romanian Mathematical Olympiad

Let $B=\{b,5b,a\}$. The possible cases are:

I. $a=\frac{b+5b}{2}\Rightarrow B=\{b,3b,5b\}$;

II. $5b=\frac{a+b}{2}\Rightarrow B=\{b,5b,9b\}$;

III. $b=\frac{a+5b}{2}\Rightarrow a+3b=0$ - impossible.

a) Since $225$ is divisible by $3$, $5$ and $9$, it can be any element of the set $B$. We get the nice sets: $\{225,675,1125\}$, $\{75,225,375\}$, $\{45,135,225\}$, $\{225,1125,2025\}$, $\{45,225,405\}$, $\{25,125,225\}$.

b) In case I we get a nice set for every $b\in {\mathbb{N}}^{\ast }$ with $5b\le 2025$, that is $2025÷5=405$ sets. In case II we get a nice set for every $b\in {\mathbb{N}}^{\ast }$ so that $9b\le 2025$, that is $2025÷9=225$ sets. Since the equality $\{x,3x,5x\}=\{y,5y,9y\}$ is impossible for positive integers $x,y$, there are no common sets for case I and case II. This shows that there are $405+225=630$ nice sets.