For each $k=1, 2, \dotsc$ , n let $S_k = \left\lfloor \frac{x+0}{k}\right\rfloor + \left\lfloor \frac{x+1}{k} \right\rfloor + \left\lfloor\frac{x+2 }{k} \right\rfloor + \dotsb + \left\lfloor \frac{x+(k-1) }{k}\right\rfloor.$
Show that $S_1+S_2+\cdots +S_n=n\lfloor x \rfloor$.
Here is what can be done:
$S_1=\lfloor x\rfloor$
$S_2=\left\lfloor\frac{x+1}{2}\right\rfloor=\left\lfloor\frac{x}{2}+\frac12\right\rfloor \rightarrow y=\frac{x}{2},\rightarrow \left\lfloor y+\frac12\right\rfloor=\left\lfloor2\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{2}\right\lfloor=\lfloor x\rfloor-\left\lfloor\frac{x}{2}\right\rfloor$
Here $S_2$ is solved by using the equality $\left\lfloor y+\frac12\right\rfloor=\lfloor 2y\rfloor -\lfloor y\rfloor$.
If we get that $S_3=\lfloor x\rfloor +\lfloor \frac{1}{2}\rfloor$ then summing it with $S_2$ would give $\lfloor2 x\rfloor$, we should be on the way to the answer $S_1+S_2+S_3=3\lfloor x\rfloor$. But which equality of floor functions can I use to find show that $\left\lfloor\frac{x+2}{3}\right\rfloor=\lfloor x\rfloor$?
Thanks