Let's integrate

$\begin{aligned} I & = \int _0^\infty \left \lfloor \frac 6{\color{#3D99F6}{e^x}} \right \rfloor dx \quad \quad \small \color{#3D99F6}{\text{We note that } \frac 6{\frac6k} = k \implies \ln \frac6k < x \le \frac 6{k-1} \implies \left \lfloor \frac 6{\color{#3D99F6}{e^x}} \right \rfloor = k} \\ & = \int _0^{\ln \frac65} 5 \ dx + \int _{\ln \frac65}^{\ln \frac64} 4 \ dx + \int _{\ln \frac64}^{\ln \frac63} 3 \ dx + \int _{\ln \frac63}^{\ln \frac62} 2 \ dx + \int _{\ln \frac62}^{\ln 6} 1 \ dx \\ & = \sum_{k=1}^5 \int_{\ln \frac6{k+1}}^{\ln \frac6k} k \ dx \\ & = \sum_{k=1}^5 k\left(\ln \frac6k - \ln \frac6{k+1} \right) \\ & = \sum_{k=1}^5 k\left(\ln (k+1) - \ln k \right) \\ & = \sum_{k= \red 2}^{\red 6} (k-1) \ln k - \sum_{k=1}^5 \ln k \\ & = \sum_{k= \red 2}^{\red 6} k \ln k - \sum_{k=1}^5 \ln k - \sum_{k=\red 2}^{\red 6} \ln k \\ & = 6 \ln 6 - \ln (6!) \\ & = \ln \frac {6^5}{5!} \end{aligned}$

$\implies A+B+C = 6+5+5 = \boxed{16}$

Generalization: $\displaystyle \int _0^\infty \left \lfloor \frac n{e^x} \right \rfloor dx = \ln \frac{n^{n-1}}{(n-1)!}$