$\log _2 \log _2 \log _2 \log _2 \log _2(n)<0<\log _2 \log _2 \log _2 \log _2(n)$. Let $l$ be the number of digits in the binary expansion of $n$. Then the minimum and the maximum possible values of $l$ are
We have,
$\log _2 \log _2 \log _2 \log _2 \log _2(n) < 0$
$ < \log _2 \log _2 \log _2 \log _2(n)$
$\begin{aligned} \log _2 \log _2 \log _2 \log _2 \log _2(n) < 0 \\ \log _2 \log _2 \log _2 \log _2(n) < 2^6 \\ \log _2 \log _2 \log _2 \log _2(n) < 1 \\ \log _2 \log _2 \log _2(n) < 2 \\ \log _2 \log _2(n) < 2^2 \\ \log _2(n) < 2^4 \\ n < 2^{16} \end{aligned}$
Similarly, for
$\log _2 \log _2 \log _2 \log _2(n) > 0 \Rightarrow n > 2^4$
Hence, $\quad 2^4 < n < 2^{16}$
$\therefore$ The minimum number of digits in binary expansion of $n$ is $5$ and maximum numbers of digits in binary expansion of $n$ is $16 .$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ $1 < e < \sqrt{2}$
$(B)$ $\sqrt{2} < e < 2$
$(C)$ $\Delta=a^4$
$(D)$ $\Delta=b^4$
