Question 81
Let $m$, $n$ and $x$ be strictly positive integers. Prove that
$$
\sum_{i=1}^{n} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, m\right)=\sum_{i=1}^{m} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, n\right)
$$
$$
\sum_{i=1}^{n} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, m\right)=\sum_{i=1}^{m} \min \left(\left\lfloor\frac{x}{i}\right\rfloor, n\right)
$$