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

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