MathDB
p(2018)= p(2019)

Source: MEMO 2018 I4

September 7, 2018
number theoryalgebrafloor function

Problem Statement

(a) Prove that for every positive integer mm there exists an integer n≥mn\ge m such that ⌊n1⌋⋅⌊n2⌋⋯⌊nm⌋=(nm) (∗)\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)p(m) the smallest integer n≥mn \geq m such that the equation (∗) (*) holds. Prove that p(2018)=p(2019).p(2018) = p(2019). Remark: For a real number x,x, we denote by ⌊x⌋\left \lfloor x \right \rfloor the largest integer not larger than x.x.
Back to ProblemsView on AoPS