Prove that
$[x]+[x+\dfrac{1}{n}]+[x+\dfrac{2}{n}]+.....[x+\dfrac{n-1}{n}]=[nx]$
Where [.] Represents greatest integer function and n$\in$N
I thought about this, too, but how do you relate $\lfloor x+\dfrac{1}{p+1}\rfloor$ to $\lfloor x+\dfrac{1}{p}\rfloor$? We know that the LHS is less than or equal to the RHS. We know it is equal to the RHS or the RHS minus one. But, without more information about $p$, we will not be able to make any assessment.
Instead, consider $\overline{x} = x-\lfloor x \rfloor$. Obviously, $0 \le \overline{x} < 1$ and $nx = n\lfloor x \rfloor + n\overline{x}$.
$\lfloor n\overline{x} \rfloor$ is the number of terms of $x, x+\dfrac{1}{n}, x+\dfrac{2}{n}, \ldots, x+\dfrac{n-1}{n}$ such that $\lfloor x+k\rfloor > \lfloor x \rfloor$ where $k \in \left\{0,\dfrac{1}{n},\dfrac{2}{n},\ldots, \dfrac{n-1}{n}\right\}$.
But, I have not thought how to prove that. Intuitively, it seems obvious, though.