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

Accepted Answer

C. We have $8^m = 2^{3m}$ and $8^n = 2^{3n}$. The integers between $2^{3m}$ and $2^{3n}$ that are powers of 2 are: $$2^{3m+1}, 2^{3m+2}, \dots, 2^{3n-1}$$ The total number of such terms is: $$(3n - 1) - (3m + 1) + 1 = 3n - 3m - 1$$ Given that there are 41 such integers: $$3n - 3m - 1 = 41 \implies 3(n - m) = 42 \implies n - m = 14$$ Since $n$ and $m$ are positive integers, the minimum value for $m$ is $1$. Then $n = m + 14 = 15$. Thus, the smallest possible value of $n + m = 15 + 1 = 16$.

Powers of 2 between Powers of 8

Explanation

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