\(\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{99}\)

\(=\left(\dfrac{1}{50}+\dfrac{1}{99}\right)+\left(\dfrac{1}{51}+\dfrac{1}{98}\right)+...+\left(\dfrac{1}{74}+\dfrac{1}{75}\right)\)

\(=\dfrac{149}{4950}+\dfrac{149}{4998}+...+\dfrac{149}{5550}=\dfrac{a}{b}\)

\(=\dfrac{149}{1}\left(\dfrac{1}{4950}+\dfrac{1}{4998}+...+\dfrac{1}{5550}\right)\)

=> a\(⋮149\)

We have : \(\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{99}=\dfrac{a}{b}\)

\(=\left(\dfrac{1}{50}+\dfrac{1}{99}\right)+\left(\dfrac{1}{51}+\dfrac{1}{98}\right)+...+\left(\dfrac{1}{74}+\dfrac{1}{75}\right)=\dfrac{a}{b}\)

\(=\dfrac{149}{50\cdot99}+\dfrac{149}{51\cdot98}+...+\dfrac{149}{74\cdot75}=\dfrac{a}{b}\)

So we have \(\dfrac{a}{b}\) kind \(\dfrac{149k}{50\cdot51\cdot...\cdot99}\)\(\Leftrightarrow\dfrac{a}{b}=\dfrac{149k}{50\cdot51\cdot...\cdot999}\Rightarrow a⋮149\left(a=149k\right)\)

Done.