CAT 2023 Slot 2 QA Question 4: 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

Prime factorise the right-hand side.

$144^{145} = (2^{4} 	imes 3^{2})^{145}$
$144^{145} = 2^{580} 	imes 3^{290}$

To maximize $n$, $b$ should be as small as possible and should use the larger available prime exponent. Taking $b = 2$ gives $n = 580$.

To minimize $m$, we can take $m = 1$ and absor

Let aaa, bbb, mmm and nnn be natural numbers such that a>1a > 1a>1 and b>1b > 1b>1. If ambn=144145a^{m}b^{n} = 144^{145}ambn=144145, then the largest possible value of n−mn - mn−m is

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