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Number of positive divisors and integer part.

Source: Tuymaada 2000, day 1, problem 1.

April 30, 2007
number theory proposednumber theory

Problem Statement

Let d(n)d(n) denote the number of positive divisors of nn and let e(n)=[2000n]e(n)=\left[2000\over n\right] for positive integer nn. Prove that d(1)+d(2)+⋯+d(2000)=e(1)+e(2)+⋯+e(2000).d(1)+d(2)+\dots+d(2000)=e(1)+e(2)+\dots+e(2000).
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