75th Romanian Mathematical Olympiad

For $x=2$, we get $2^k=|f_1(2)|\cdot |f_2(2)|\cdot \cdots \cdot |f_n(2)|$. Using injectivity, we deduce that $|f_i(2)|\ge 2$ for any $i=\overline{1,n}$. Thus, $2^k=|f_1(2)|\cdot |f_2(2)|\cdot \cdots \cdot |f_n(2)|\ge 2^n$, so $k\ge n$. Since $n$ and $k$ have the same parity, it follows that $M_k\subset \{k,k-2,k-4,\dots \}\subset {\mathbb{N}}^{\ast }$. We will prove that all values of the form $n=k-2t$ are good. For this, we observe that $x^k=\underset{k-2t-1\text{ times}}{\underbrace{x\cdot x\cdots x}}\cdot x^{2t+1}$, so functions $f_i(x)=x$, $i=\overline{1,n-1}$, $f_n(x)=x^{2t+1}$ are injective and $x^n=f_1(x)\cdot f_2(x)\cdots f_{n-1}(x)\cdot f_n(x)$. In conclusion, $k-2t\in M_k$ for any $t\in \mathbb{N}$ with $2t<k$ and $M_k=\{k,k-2,k-4,\dots ,k-2\cdot \lfloor \frac{k-1}{2}\rfloor \}$.