Let n be a positive integer such that _2 _2 _2 _2 _2(n)<0< _2 _2 … - Vidyadip

MCQ

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

✓
$5$ and $16$
B
$5$ and $17$
C
$4$ and $16$
D
$4$ and $17$

✓

Answer

Correct option: A.

$5$ and $16$

a
(a)

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 $16 .$

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