AMC8 2022 8, Problem 8

What is the value of \[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}?\]

Solution: Note that common factors (from $3$ to $20,$ inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes \[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\fbox{\frac{1}{231}}.\]

Answer: \frac{1}{231}