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CAT 2023 Slot 3 QA Question & Solution

AlgebraMedium

Question

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

Options

42
44
14
16

Solution

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $8^m$ and $8^n$, (where m, and n are positive integers, and m < n)

Hence, $2^{3m}\ <\ \text{41 numbers }<\ 2^{3n}$

Since, m is a positive integer, the least value of m is 1. Therefore, $2^{3m\ }=2^3$, hence, the 41 numbers between them are $2^4,\ 2^5,\ 2^6,...,2^{44}$.

Then the lowest possible value of $8^n$ is $2^{45}$. Hence, the smallest value of n is $2^{45}\ =\ 8^n\ =>\ 2^{3n\ }=2^{45\ }=>\ n\ =\ 15$

Hence, the smallest value of m+n is (15+1) = 16

The correct option is D