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:

Question

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:

Accepted Answer

To solve the problem, we start by factorizing $$144^{145}$$:  
$$
144 = 2^4 	imes 3^2 \quad 	ext{so} \quad 144^{145} = (2^4 	imes 3^2)^{145} = 2^{580} 	imes 3^{290}.
$$

We are given $$a^m b^n = 144^{145}$$. To maximize $$n - m$$, we need to assign the exponents $$m$$ and $$n$$ to minimize $$m$$ while keeping the values valid.

- Let $$a = 3^{290}$$, which implies $$m = 1$$.  
- Let $$b = 2^{580}$$, which implies $$n = 580$$.

Thus, the difference between $$n$$ and $$m$$ is:  
$$
n - m = 580 - 1 = 579.
$$

This configuration satisfies all conditions ($$a > 1, b > 1, m, n \in \mathbb{N}$$), and hence the largest possible value of $$n - m$$ is **579**.

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