CAT 2023 Slot 2 Solved Paper

It is given that ${\mathit{a}}^{\mathit{m}}\cdot {\mathit{b}}^{\mathit{n}}={144}^{145}$, where $\mathit{a}>1$ and $\mathit{b}>1$.
144 can be written as $144=2^4\times 3^2$
Hence, ${\mathit{a}}^{\mathit{m}}\cdot {\mathit{b}}^{\mathit{n}}={144}^{145}$ can be written as ${\mathit{a}}^{\mathit{m}}\cdot {\mathit{b}}^{\mathit{n}}={\left(2^4\times 3^2\right)}^{145}=2^{580}\times 3^{290}$
We know that $3^{290}$ is a natural number, which implies it can be written as ${\mathit{a}}^1$, where $\mathit{a}>1$
Hence, the least possible value of $\mathit{m}$ is 1 . Similarly, the largest value of $\mathit{n}$ is 580 .
Hence, the largest value of $\left(\mathit{n}-\mathit{m}\right)$ is $\left(580-1\right)=579$
The correct option is D