Let's do some calculus! (26)

The problem can be solved using Riemann sums as follows.

$\begin{aligned} S & = \lim_{n \to \infty} n^{-\frac 32} \sum_{j=1}^{6n} \sqrt j = \lim_{n \to \infty} \frac 1n \sum_{j=1}^{6n} \sqrt{\frac jn} = \int_a^b x^\frac 12 dx \end{aligned}$

Lower limit $a = \displaystyle \lim_{n \to \infty} \frac 1n = 0$ and upper limit $b = \displaystyle \lim_{n \to \infty} \frac {6n}n = 6$ . Therefore, we have:

$\begin{aligned} S & = \int_0^6 x^\frac 12 dx = \frac 23 x^\frac 32 \bigg|_0^6 \approx 9.798 \end{aligned}$

$\implies \lfloor S \rfloor = \boxed{9}$