CAT 2023 Slot 3 QA Q2 Logarithms & Exponents

Question 2 of the Quantitative Ability (QA) section, CAT 2023 Slot 3, with CAT's official answer and a worked solution.

Let \(n\) and \(m\) be two positive integers such that there are exactly 41 integers greater than \(8^m\) and less than \(8^n\), which can be expressed as powers of 2 . Then, the smallest possible value of \(n+m\) is

A
14
B
42
C
16
D
44

Answer: (C) 16 Logarithms & Exponents MCQ

Solution

\(8^m = 2^{3m}\) and \(8^n = 2^{3n}\). Powers of 2 strictly between them are \(2^{3m+1},\dots,2^{3n-1}\): that is \(3n-3m-1\) numbers.
\(3(n-m)-1 = 41 \Rightarrow n-m = 14\).
Smallest \(n+m\) takes \(m=1, n=15\): \(n+m=16\).