Removing Floors - Part II

First note that $\dfrac n2 + \dfrac n3 + \dfrac n6 =n$ . So we have, $\left \lfloor \dfrac n2 \right \rfloor + \left \lfloor \dfrac n3 \right \rfloor + \left \lfloor \dfrac n6 \right \rfloor = \dfrac n2 + \dfrac n3 + \dfrac n6$

Since $\{x\}= x -\lfloor x \rfloor.$ we have $\{\dfrac n2\} + \{\dfrac n3\} + \{\dfrac n6 \} =0$

Since $\{x\}\ \geq 0$ it follows that $\dfrac n2, \dfrac n3, \dfrac n6$ are all natural numbers. Thus $n$ is divisible by $6$ . Also all the numbers which are divisible by $6$ satisfy the equation. There are $16$ multiples of $6$ in the given interval which is why the answer is $\boxed { 16 }$ .

$\Bigl\lfloor\dfrac{n}{2}\Bigr\rfloor + \Bigl\lfloor\dfrac{n}{3}\Bigr\rfloor + \Bigl\lfloor\dfrac{n}{6}\Bigr\rfloor = \dfrac{n}{2} + \dfrac{n}{3} +\dfrac{n}{6} \\ \implies (\dfrac{n}{2} - \Bigl\lfloor\dfrac{n}{2}\Bigr\rfloor) + (\dfrac{n}{3} - \Bigl\lfloor\dfrac{n}{3}\Bigr\rfloor) + (\dfrac{n}{6} - \Bigl\lfloor\dfrac{n}{6}\Bigr\rfloor) = 0 \\ \implies frac(\frac{n}{2 }) + frac(\frac{n}{3}) + frac(\frac{n}{6}) = 0 \\$
The fractional part function is $\geq$ 0. Since the sum of the terms is 0, each of them must be 0. Thus $\dfrac{n}{2}$ , $\dfrac{n}{3}$ and $\dfrac{n}{6}$ must be integers.Thus, the question reduces to find the number of multiples of 6 from 1 to 100, which is $\boxed{16}$

The only case

$n= 6\kappa$ .

Such that $\kappa \in [1,16]$ .

$\kappa \in N$ Hence the conclusion follows.