AMC8 2022

AMC8 2022 · Q8

AMC8 2022 · Q8. It mainly tests Fractions.

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

计算\[\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdots\frac{18}{20}\cdot\frac{19}{21}\cdot\frac{20}{22}\]的值。

(A) \frac{1}{462} \frac{1}{462}

(B) \frac{1}{231} \frac{1}{231}

(D) \frac{2}{213} \frac{2}{213}

(E) \frac{1}{22} \frac{1}{22}

Answer

Correct choice: (B)

正确答案：（B）

Solution

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

注意分子和分母的公因数（从$3$到$20$，包括两端）可以消去。因此，原表达式简化为\[\frac{1}{\cancel{3}}\cdot\frac{2}{\cancel{4}}\cdot\frac{\cancel{3}}{\cancel{5}}\cdots\frac{\cancel{18}}{\cancel{20}}\cdot\frac{\cancel{19}}{21}\cdot\frac{\cancel{20}}{22}=\frac{1\cdot2}{21\cdot22}=\frac{1}{21\cdot11}=\boxed{\textbf{(B) } \frac{1}{231}}.\]

Topics

AR.001 Fractions