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RO-TST 2012 Problem 6

    Prove that for any positive integer n≥2n\geq 2 we have that ∑k=2n⌊nk⌋=∑k=2n⌊log⁡kn⌋.\sum_{k=2}^n \lfloor \sqrt[k]{n}\rfloor=\sum_{k=2}^n\lfloor\log_{k}n\rfloor.

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