MCQ
The number $111......1 $ ( $ 91$ times) is
- ✓Not a prime
- BAn even number
- CNot an odd number
- DNone of these
= $1 + 10 + {10^2} + ..... + {10^{90}}$
= $\frac{{{{10}^{91}} - 1}}{{10 - 1}} = \frac{{{{({{10}^7})}^{13}} - 1}}{{10 - 1}}$= $\frac{{{t^{13}} - 1}}{9}$, where $t = {10^7}$
= $\left( {\frac{{t - 1}}{9}} \right)\,({t^{12}} + {t^{11}} + ..... + t + 1)$
= $\left( {\frac{{{{10}^7} - 1}}{{10 - 1}}} \right)\,(1 + t + {t^2} + .... + {t^{12}})$
$ = (1 + 10 + {10^2} + .... + {10^6})(1 + t + {t^2} + ... + {t^{12}})$
$111.....1(91\,\,{\rm{times)}}$ is a composite number.
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$\log _2 \log _2 \log _2 \log _2 \log _2(n)<0<\log _2 \log _2 \log _2 \log _2(n)$. Let $l$ be the number of digits in the binary expansion of $n$. Then the minimum and the maximum possible values of $l$ are