Question 97
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 $x$.
$$
\left\lfloor\frac{n}{1}\right\rfloor \cdot\left\lfloor\frac{n}{2}\right\rfloor \cdots\left\lfloor\frac{n}{m}\right\rfloor=\binom{n}{m} \tag{*}
$$
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 $x$.