2022 AMC 8 第 8 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

8.

下列表达式的值是多少?

13⋅24⋅35⋯1820⋅1921⋅2022 \dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}

What is the value of:

13⋅24⋅35⋯1820⋅1921⋅2022 \dfrac{1}{3} \cdot \dfrac{2}{4} \cdot \dfrac{3}{5} \cdots \dfrac{18}{20} \cdot \dfrac{19}{21} \cdot \dfrac{20}{22}

1462\displaystyle \dfrac{1}{462}

1231\displaystyle \dfrac{1}{231}

1132\displaystyle \dfrac{1}{132}

2213\displaystyle \dfrac{2}{213}

122\displaystyle \dfrac{1}{22}

答案:B
知识点:裂项相消分数
难度评级:1020
小提示:

大多数分子和分母因子会相消

Most numerator and denominator factors cancel

大提示:

相消后,分子只剩 1⋅21\cdot2

After cancellation, only 1⋅21\cdot2 remains on top

解答:

从 33 到 2020 的每个整数都既作为分母出现一次,也作为分子出现一次,所以全部相消。

相消后,分子只剩 11 和 22,分母只剩 2121 和 2222。

剩下的分数是 1⋅221⋅22\dfrac{1 \cdot 2}{21 \cdot 22} ,化简为 2462=1231\dfrac{2}{462} = \dfrac{1}{231}。

正确答案是 B。

Since every integer from 33 to 2020 occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only 11 and 22 left as numerators and 2121 and 2222 left as denominators.

The remaining fraction is 1⋅221⋅22. \dfrac{1 \cdot 2}{21 \cdot 22} . This simplifies to 2462=1231 \dfrac{2}{462} = \dfrac{1}{231}

Thus, the correct answer is B.

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