Solusi

Misalkan $X=\frac{1}{1980}+\frac{1}{1981}+...+\frac{1}{2001}$

Batas bawah

$\frac{1}{1980}+\frac{1}{1981}+...+\frac{1}{2001}<\frac{1}{1980}+\frac{1}{1980}+...+\frac{1}{1980}$

$X<\frac{1}{1980}+\frac{1}{1980}+...+\frac{1}{1980}$

$X<\frac{22}{1980}$

$X<\frac{1}{90}$

$A>90$

Batas atas

$\frac{1}{1980}+\frac{1}{1981}+...+\frac{1}{2001}>\frac{1}{2001}+\frac{1}{2001}+...+\frac{1}{2001}$

$X>\frac{1}{2001}+\frac{1}{2001}+...+\frac{1}{2001}$

$X>\frac{22}{2001}$

$A<\frac{2001}{22}$

$A<90\frac{21}{22}$

Dari batas atas dan batas bawah kita dapat $90<A<90\frac{21}{22}$

Cukup jelas $\lfloor A\rfloor =90$