MEMO 2018 I4

(a) Prove that for every positive integer $m$ there exists an integer $n\ge m$ such that $$\left \lfloor \frac{n}{1} \right \rfloor \cdot \left \lfloor \frac{n}{2} \right \rfloor \cdots \left \lfloor \frac{n}{m} \right \rfloor =\binom{n}{m} \ (*)$$ (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 $\left \lfloor x \right \rfloor$ the largest integer not larger than $x.$