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

We are given that $n$ and $m$ are positive integers. We want to find the number of integers that are strictly between $8^m$ and $8^n$ and can be expressed as powers of 2.

First, let us express the boundaries as powers of 2:
$8^m={\left(2^3\right)}^m=2^{3m}$
$8^n={\left(2^3\right)}^n=2^{3n}$

Let the integers that are powers of 2 be represented as $2^k$, where $k$ is an integer. The condition that these integers lie strictly between $8^m$ and $8^n$ is written as:
$2^{3m}<2^k<2^{3n}$

Since the exponential function with base 2 is strictly increasing, this inequality simplifies to a relation between the exponents:
$3m<k<3n$

Since $k$ must be an integer, the possible values for $k$ are the integers in the range:
$k\in \left\{3m+1,3m+2,...,3n-1\right\}$

The total count of these integers is given by:
$\left(3n-1\right)-\left(3m+1\right)+1=3n-3m-1$

We are given that there are exactly 41 such integers. Therefore, we can set up the equation:
$3n-3m-1=41$
$3\left(n-m\right)=42$
$n-m=14$

From this, we express $n$ in terms of $m$:
$n=m+14$

We want to find the smallest possible value of $n+m$. Substituting $n$ into the expression gives:
$n+m=\left(m+14\right)+m=2m+14$

Since $m$ must be a positive integer, the smallest possible value it can take is:
$m=1$

Substituting $m=1$ back into the expression:
$n+m=2\left(1\right)+14=16$

Thus, the smallest possible value of $n+m$ is 16.