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

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