What's that time?

Let $\textbf{H : M}$ be the time with that property between $\textbf{H}$ and $\textbf{H + 1}$ where $0 \leq H \leq 11 , \textbf{H}\in \mathbb{Z}$ and $\textbf{M} \in \mathbb{R}$

We know that in 1 minute, minute hand rotate through $6^{\circ}$ and hour hand rotate through $\frac{1}{2}^{\circ}$ .

So, the angular position of minute hand with respect to reference line (line joining centre to 12) in clockwise direction will be $6M^{\circ}$ and that of hour hand in clockwise direction will be ${30H}^{\circ} + \frac{M}{2}^{\circ}$ .

Let after interchanging the hands, the time be $H_{R} : M_{R}$ here also $H_{R}$ and $M_{R}$ are defined as $H$ and $M$ .

Now, the minute hand will be at the angular position of ${30H}^{\circ} + \frac{M}{2}^{\circ}$ and hour hand will be at the angular position of $6M^{\circ}$

i.e. $6M_{R}^{\circ} = {30H}^{\circ} + \frac{M}{2}^{\circ}$ $\cdots Eq. 1$ and $30H_{R}^{\circ} + \frac{M_{R}}{2}^{\circ} = 6M^{\circ}$ $\cdots Eq. 2$

Using Eq. 1 in Eq. 2 ,

$30H_{R}^{\circ} = \LARGE 6M^{\circ} - \frac{\frac{30H^{\circ} + \frac{M}{2}^{\circ}}{6}}{2}$

$\Rightarrow 30H_{R} = \frac{143M}{24} -\frac{5H}{2}$ $\cdots Eq. 3$

But $H_{R}$ is a integers, so, $H_{R}$ is greatest integer less than or equal to $\frac{6M^{\circ}}{30^{\circ}} \\ \Rightarrow H_{R} = \bigg\lfloor\frac{6M^{\circ}}{30^{\circ}}\bigg\rfloor \cdots Eq. 4$

Using Eq. 4 in Eq. 3

$30\bigg\lfloor\frac{6M^{\circ}}{30^{\circ}}\bigg\rfloor = \frac{143M}{24} -\frac{5H}{2} \\ \Rightarrow \frac{143M}{24} = \frac{5H}{2} + 30n$

where $n = \bigg\lfloor\frac{6M}{30}\bigg\rfloor = \bigg\lfloor\frac{M}{5}\bigg\rfloor \\ Now , \\ 0 \leq M < 60 \\ \Rightarrow 0 \leq \frac{M}{5} <12 \\ \Rightarrow 0 \leq \bigg\lfloor\frac{M}{5}\bigg\rfloor \leq 11 \\ \Rightarrow 0 \leq n \leq 11 \quad, n \in \mathbb{Z}$

$\Rightarrow M = \frac{24}{143}\times(\frac{5H}{2} + 30n)$

$\Rightarrow M = \large\frac{60(H + 12n)}{143}$

So the time with that property in between $H$ and $H + 1$ is

$\large H : \frac{60(H + 12n)}{143} \quad , 0 \leq n \leq 11$

We see that for every permissible value of $H$ there are $12$ values of $n$ and so $12$ values of $M$ . And there are 12 possible values of $H$ form $0$ to $11$ . So in 12 hours there are $12\times12 = 144$ times possible with that property. But in counting time 12 :00 is counted 2 times , one by taking H = n = 11 and other by taking H = n = 0.

Hence there are 144 - 1 = 143 times possible . $\Rightarrow m = 143$

Comparing the calculated expression of $M$ with $\large\frac{a(H + bn)}{c}$ we get

$a = 60\quad, b = 12\quad, c = 143$

$\LARGE \sqrt{\bigg\lfloor\frac{c}{b}\bigg\rfloor - \bigg\lfloor\frac{m}{a}\bigg\rfloor} = \LARGE \sqrt{\bigg\lfloor\frac{143}{12}\bigg\rfloor - \bigg\lfloor\frac{143}{60}\bigg\rfloor} = 3$