Floor Function-1

Given that $\displaystyle \sum_{k=1}^{10} \left \lfloor \frac n{k!} \right \rfloor = 2019$ , we can estimate $n \approx \dfrac {2019}{e-1} \approx 1175.011$ . Obviously $n \ge 1176$ . Since $7! = 5040$ , we need only consider $k$ from $1$ to $6$ , because $\left \lfloor \dfrac n{k!} \right \rfloor = 0$ for $k \ge 7$ .

$\begin{aligned} \sum_{k=1}^6 \left \lfloor \frac {1176}{k!} \right \rfloor & = \left \lfloor \frac {1176}1 \right \rfloor + \left \lfloor \frac {1176}2 \right \rfloor + \left \lfloor \frac {1176}6 \right \rfloor + \left \lfloor \frac {1176}{24} \right \rfloor + \left \lfloor \frac {1176}{120} \right \rfloor + \left \lfloor \frac {1176}{720} \right \rfloor \\ & = 1176 + 588 + 196+49+9+1 = 2019 \end{aligned}$

If $n=1177$ , the first term will be $1177$ and the sum will be $2020$ . Therefore $n=1176$ is the only solution and the sum of $n$ is $\boxed{1176}$ .