CAT 2023 Slot 2 QA Q1 Logarithms & Exponents

Question 1 of the Quantitative Ability (QA) section, CAT 2023 Slot 2, with CAT's official answer and a worked solution.

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

A
580
B
290
C
579
D
289

Answer: (C) 579 Logarithms & Exponents MCQ

Solution

\(144^{145} = (2^4\cdot 3^2)^{145} = 2^{580}\cdot 3^{290}\).
To make \(n\) large and \(m\) small, take \(m = 1\) and let \(b^n\) absorb as many prime factors as possible.
With \(b = 2\), \(n = 580\) and \(a = 3^{290} > 1\), so \(a^1 b^{580} = 144^{145}\) works.
\(n\) cannot exceed 580 (the largest exponent available for any \(b>1\)), and \(m \ge 1\), so the largest \(n - m = 580 - 1 = 579\).

Shortcut. Max \(n\) is the largest prime exponent (580), min \(m\) is 1, so the answer is 579.