Let n be a positive integer such that log _2 | vedclass

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(A) Given the inequality: $\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 < \log _2 \log _2 \log _2 \log _2(n)$.First,consider $\log _2 \log _2 \log _

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$5$ and $16$

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(A) Given the inequality: $\log _2 \log _2 \log _2 \log _2 \log _2(n) First,consider $\log _2 \log _2 \log _2 \log _2(n) > 0$:$\log _2 \log _2 \log _2 \log _2(n) > 0 \implies \log _2 \log _2 \log _2(n) > 1 \implies \log _2 \log _2(n) > 2 \implies \log _2(n) > 4 \implies n > 2^4 = 16$.Next,consider $\log _2 \log _2 \log _2 \log _2 \log _2(n) $\log _2 \log _2 \log _2 \log _2 \log _2(n) Thus,$16 The number of digits $l$ in the binary expansion of $n$ is given by $\lfloor \log_2(n) \rfloor + 1$.For $n > 16$,the minimum value is $\lfloor \log_2(16 + 1) \rfloor + 1 = 4 + 1 = 5$.For $n Therefore,the minimum and maximum values of $l$ are $5$ and $16$.

Let $n$ be a positive integer such that $\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 < \log _2 \log _2 \log _2 \log _2(n)$. Let $l$ be the number of digits in the binary expansion of $n$. Then the minimum and the maximum possible values of $l$ are

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