2022 AMC 8 Problem 8

Attempt Problem 8 of the 2022 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

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}
Answer: B
Concepts:telescopingfraction
Small hint:
Most numerator and denominator factors cancel
Big hint:
After cancellation, only 1⋅21\cdot2 remains on top
Solution:
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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