Question

Let $[k]$ denotes the greatest integer less than or equal to $k$. If number of positive integral solutions of the equation $\left[\frac{x}{\left[{\pi }^2\right]}\right]=\left[\frac{x}{\left[11\frac{1}{2}\right]}\right]$ is $n$,

Answer 1

Answer: 4

$\left[\frac{x}{9}\right]=\left[\frac{x}{11}\right]$
Case-I : $0\le \frac{x}{9}<1$ and $0\le \frac{x}{11}<1$
$\Rightarrow 0\le x<9$ and $0\le x<11\Rightarrow$ common value of $x$ is $\{1,2,3,\dots ..8\}$
Case-II : $1\le \frac{x}{9}<2$ and $1\le \frac{x}{11}<2$
$\Rightarrow 9\le x<18$ and $11\le x<22\Rightarrow x\in \{11,12,\dots ..,17\}$
Case-III : $2\le \frac{x}{9}<3$ and $2\le \frac{x}{11}<3$
$\Rightarrow 18\le x<27$ and $22\le x<33\Rightarrow x\in \{22,23,\dots \dots ,26\}$
Case-IV : $3\le \frac{x}{9}<4$ and $3\le \frac{x}{11}<4$
$\Rightarrow 27\le x<36\text{ and }33\le x<44\Rightarrow x\in \{33,34,35\}$
Case-V : $4\le \frac{x}{9}<5$ and $4\le \frac{x}{11}<5\Rightarrow x=44$
$\therefore$ total positive integer $x=8+7+5+3+1=24$
$\therefore$ Answer $=\sqrt{24-8}=8$ .