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:

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Accepted Answer

To determine the smallest possible value of $$ n + m $$, follow these steps:

1. **Understand the problem**: We need to find two integers $$ n $$ and $$ m $$ such that there are exactly 41 integers between $$ 8^m $$ and $$ 8^n $$, which can be expressed as powers of 2.

2. **Express powers of 2 within the range**: The integers between $$ 8^m $$ and $$ 8^n $$ will be of the form $$ 2^{3m}, 2^{3m+1}, \dots, 2^{3n-1} $$. The total count is given by:
   $$
   (3n - 1) - (3m) + 1 = 3n - 3m + 1.
   $$
   Simplify this to:
   $$
   3n - 3m = 41.
   $$

3. **Solve for $$ n - m $$**: Dividing through by 3 gives:
   $$
   n - m = 14.
   $$

4. **Minimize $$ n + m $$**: To minimize $$ n + m $$, choose the smallest possible values for $$ m $$. Let $$ m = 1 $$, then:
   $$
   n = m + 14 = 15.
   $$
   Thus, the smallest value of $$ n + m = 1 + 15 = 16.

Hence, the correct answer is **16**.

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