CAT 2019 Shift 2 Paper

Given, $\mathit{f}\left(\mathit{m}\mathit{n}\right)=\mathit{f}\left(\mathit{m}\right)\mathit{f}\left(\mathit{m}\right)$
Also, $\mathit{f}\left(24\right)=54$
$\Rightarrow \mathit{f}\left(24\right)=2\times 3\times 3\times 3$
⇒ f(2 × 12)=f(2) f(12)=f(2) f(2 × 6)=f(2) f(2) f(6) = f(2) f(2) f(2 × 3) = f(2) f(2) f(2) f(3) = 2 × 3 × 3 × 3
Given that $\mathit{f}\left(1\right),\mathit{f}\left(2\right)$ and $\mathit{f}\left(3\right)$ are all positive integers, by comparison, we get
$\mathit{f}\left(2\right)=3$ and $\mathit{f}\left(3\right)=2$. And we can safely take $\mathit{f}\left(1\right)=1$
Now, f(18)=f(2)(9)=f(2) f(3 × 3)=f(2) f(3) f(3)=3 × 2 × 2 = 12