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

We need to find two positive integers $n$ and $m$ such that exactly 41 powers of 2 lie between $8^m$ and $8^n$.

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Since we're dealing with powers of 2, let's convert $8^m$ and $8^n$ to base 2:

$8^m = (2^3)^m = 2^{3m}$

$8^n = (2^3)^n = 2^{3n}$

Since 8 = 2³, this makes it easier to work with powers of 2.

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The powers of 2 that are:

Greater than $8^m = 2^{3m}$

Less than $8^n = 2^{3n}$

would be: $2^{3m+1}, 2^{3m+2}, 2^{3m+3}, ..., 2^{3n-1}$

The smallest power of 2 greater than $2^{3m}$ is $2^{3m+1}$, and the largest power of 2 less than $2^{3n}$ is $2^{3n-1}$.

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The number of powers from $2^{3m+1}$ to $2^{3n-1}$ is:

$(3n-1) - (3m+1) + 1 = 3n - 3m - 1$

When counting consecutive integers from $a$ to $b$, the count is $b - a + 1$.

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Since there are exactly 41 such powers:

$3n - 3m - 1 = 41$

$3n - 3m = 42$

$3(n - m) = 42$

$n - m = 14$

Therefore: $n = m + 14$

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To find the smallest possible value of $n + m$:

$n + m = (m + 14) + m = 2m + 14$

Since $m$ must be a positive integer, the smallest value is $m = 1$.

Therefore:

$m = 1$

$n = 1 + 14 = 15$

$n + m = 16$

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The smallest possible value of $n + m$ is $16$.

When dealing with problems involving powers with the same base, convert everything to that common base first. This makes counting and arithmetic much cleaner.