<s>[INST] Evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $ For any $x \in \mathbb{R}$  and $m \in \mathbb{N} $ evaluate $\lfloor \frac{x}{m}  \rfloor + \lfloor  \frac{x+1}{m} \rfloor  + \dots + \lfloor \frac{x+m-1}{m} \rfloor $.
Well if $x=m$ then we obviously have $\lfloor \frac{x}{m}  \rfloor + \lfloor  \frac{x+1}{m} \rfloor  + \dots + \lfloor \frac{x+m-1}{m} \rfloor = 1 + 1 + \dots + 1=m$.
If $x=-m$ then $\lfloor \frac{x}{m}  \rfloor + \lfloor  \frac{x+1}{m} \rfloor  + \dots + \lfloor \frac{x+m-1}{m} \rfloor=-m $.
If $x=0$ then $\lfloor \frac{x}{m}  \rfloor + \lfloor  \frac{x+1}{m} \rfloor  + \dots + \lfloor \frac{x+m-1}{m} \rfloor=0 $.
There are many such cases I can think of, but something tells me there might be an easier solution. 
 [/INST] This is known as Hermite's identity.  The result will always be $\lfloor x \rfloor$.
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