AoPS - Problem

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Problem

Source: Tuymaada 2000, day 1, problem 1.

Tags: number theory proposed, number theory

mathmanman

30.04.2007 22:26

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 $d(1)+d(2)+\dots+d(2000)=e(1)+e(2)+\dots+e(2000).$

Kouichi Nakagawa

30.04.2007 23:21

$d(n)$ mean　the number of lattice points in

$xy=n$ .

$e(n)$ mean the number that counted these lattice points among lengthwise direction.

N.T.TUAN

03.05.2007 20:02

$\sum_{k=1}^{2000}d(k)=\sum_{k=1}^{2000}\sum_{d|k}1=\\ \sum_{d=1}^{2000}\sum_{d|k,k\in\{1,2,...,2000\}}1=\sum_{d=1}^{2000}[\frac{2000}{d}]=\sum_{k=1}^{2000}e(k)$ .

Max D.R.

29.10.2009 20:53

In general: $d(1)+d(2)+\dots+d(n)=e(1)+e(2)+\dots+e(n)$ . Using this equality we can solve the problem 3 of the Shortlist IMO 2006.