Black book 4

$\sqrt{\left \lfloor x + \left \lfloor \dfrac{x}{2} \right \rfloor \right \rfloor} + \left \lfloor \sqrt{ \{x\}} + \left \lfloor \dfrac{x}{3} \right \rfloor \right \rfloor = 3$

Let $\begin{cases} A = \sqrt{\left \lfloor x + \left \lfloor \dfrac{x}{2} \right \rfloor \right \rfloor} \\ B = \left \lfloor \sqrt{ \{x\}} + \left \lfloor \dfrac{x}{3} \right \rfloor \right \rfloor \end{cases}$

• From $A$ , we know that $x>0$
• Since the RHS and $B$ are integers, $A$ must also be an integer.
• We also note that $B = 0$ , for $x < 3$ , $A \le 2$ and $A+B \le 2$ , therefore the solution $x \ge 3$ .
• When $x \in [3, 4)$ , we note that $A=\sqrt{4} = 2$ , $B = 1$ and $A+B = 3$

Therefore, the solution $x \in [3,4)$ and $ab = 3\times 4 = \boxed{12}$ .

The fractional part is of no use and makes no.sense. The remaining expression either the square root becomes 1 and x/3 becomes 2 -not possible. Or the sqaure rooted expression takes the value 2 and x/3 is 1.which is possible at only x=3 and all values till 4 .