Middle European Mathematical Olympiad 2018 Problem I-4

(a) Prove that for every positive integer $m$ there exists an integer $n \geq m$ such that

$$\left\lfloor \frac {n}{1} \right\rfloor \cdot \left\lfloor \frac {n}{2} \right\rfloor \cdots \left\lfloor \frac {n}{m} \right\rfloor = \binom {n} {m}. \tag{*}$$

(b) Denote by $p(m)$ the smallest integer $n \geq m$ such that the equation (*) holds. Prove that $p(2018) = p(2019)$.

Remark: For a real number $x$, we denote by $\lfloor x \rfloor$ the largest integer not larger than $x$.