Let $d(n)$ denote the number of positive divisors of $n$ and let $e(n)=\left[2000\over n\right]$ for positive integer $n$. Prove that \[d(1)+d(2)+\dots+d(2000)=e(1)+e(2)+\dots+e(2000).\]
Source: Tuymaada 2000, day 1, problem 1.
Tags: number theory proposed, number theory
Let $d(n)$ denote the number of positive divisors of $n$ and let $e(n)=\left[2000\over n\right]$ for positive integer $n$. Prove that \[d(1)+d(2)+\dots+d(2000)=e(1)+e(2)+\dots+e(2000).\]