Let a, b, m and n be natural numbers such that $$a>1$$ and $$b>1$$. If $$a^{m}b^{n}=144^{145}$$, then the largest possible value of $$n-m$$ is

It is given that $$a^m\cdot b^n\ =\ 144^{145}$$, where $$a>1\ $$ and $$b>1$$.

$$144$$ can be written as $$144\ =\ 2^4\times\ 3^2$$

Hence, $$a^m\cdot b^n\ =\ 144^{145}$$ can be written as $$a^m\cdot b^n\ =\ \left(2^4\times\ 3^2\right)^{^{145}}=2^{580}\times\ 3^{290}$$

We know that $$3^{290}$$ is a natural number, which implies it can be written as $$a^1$$, where $$a\ >\ 1$$

Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.

Hence, the largest value of (n-m) is (580-1) = 579

The correct option is D