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a) Prove that for very positive integer m there exists an integer n \geq m such that \bigg\lfloor\frac{n}{1}\bigg\rfloor\cdot\bigg\lfloor\frac{n}{2}\bigg\rfloor\cdot\ldots\cdot\bigg\lfloor\frac{n}{m}\bigg\rfloor = \binom{n}{m}

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

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