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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.

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this fucntion is periodic $ f(x)=f(x+m) $ so you could try a Fourier series representation for it – Jose Garcia Oct 11 '13 at 16:06
up vote 2 down vote accepted

This is known as Hermite's identity. The result will always be $\lfloor x \rfloor$.

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