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

• A. 44
• B. 16
• C. 42
• D. 14

Answer 1

Answer: (B)

16

Answer 2

41 integers between 8m and 8n that can be represented as powers of two are required.
That is, 41 integers that fall between 23m and 23n and can be represented as powers of two are required.
These will be the numbers: $2^{3m}, 2^{3n+1}, 2^{3m+2}, 2^{3m+3}, \ldots, 2^{3m+41}, 2^{3n}$
Obviously, $3n−1=3m+41$
$3(n-m) = 42$
$n - m = 14$
If m can only take a value of 1, then n = 15.
m + n = 1 + 15 = 16
The correct option is (B): 16.