# CAT 2021 Set-3 | Quantitative Aptitude | Question: 18

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If $n$ is a positive integer such that $( \sqrt[7]{10}) ( \sqrt[7]{10})^{2} \dots ( \sqrt[7]{10})^{n} > 999,$ then the smallest value of $n$ is

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Given that,

$(\sqrt[7]{10})(\sqrt[7]{10})^{2} \dots (\sqrt[7]{10})^{n} > 999 \; ; n \in \mathbb{Z}^{+}$

$\Rightarrow 10^{\frac{1}{7}} \times 10^{\frac{2}{7}} \times 10^{\frac{3}{7}} \times \dots \times 10^{\frac{n}{7}} >999$

$\Rightarrow 10^\left({\frac{1}{7}+\frac{2}{7}+\frac{3}{7}+ \dots + \frac{n}{7}}\right) > 999$

$\Rightarrow 10^{\frac{1}{7}(1+2+3+ \dots +n)} > 999$

$\Rightarrow 10^{\frac{1}{7}\left[\frac{n(n+1)}{2}\right]} > 999$

$\Rightarrow 10^{\frac{n(n+1)}{14}} > 10^{3}-1$

Here, $\frac{n(n+1)}{14} \geq 3$

$\Rightarrow n(n+1) \geq 42$

$\Rightarrow n^{2}+n-42 \geq 0$

$\Rightarrow n^{2}+7n-6n-42 \geq 0$

$\Rightarrow n (n+7)-6(n+7) \geq 0$

$\Rightarrow (n+7)(n-6) \geq 0$

$\Rightarrow n \geq 6\;\text{(or)} \; {\color{Red}{n \geq -7\;\text{(rejected)}} }$

$\Rightarrow \boxed{n \geq 6}$

$\therefore$ The smallest value of $n$ is $6.$

Correct Answer $:6$
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