1985 AMC 8 Problem 9

Attempt Problem 9 of the 1985 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 1985 AMC 8 solutions, or check the answer key.

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

9.

What is the value of the product of the 99 factors below?
(1−12)(1−13)(1−14)⋯⋅(1−110) \begin{aligned} &\left(1 - \tfrac12\right)\left(1 - \tfrac13\right)\left(1 - \tfrac14\right) \cdots \\ &\quad {}\cdot \left(1 - \tfrac{1}{10}\right) \end{aligned}
110\dfrac{1}{10}
19\dfrac{1}{9}
12\dfrac{1}{2}
1011\dfrac{10}{11}
112\dfrac{11}{2}
Answer: A
Concepts:telescopingfraction
Small hint:
Rewrite each factor: 1−1k=k−1k1 - \dfrac1k = \dfrac{k-1}{k}
Big hint:
Write the product as 12⋅23⋅34⋯910\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10} and watch consecutive terms cancel
Solution:
Each factor 1−1k1 - \dfrac1k equals k−1k,\dfrac{k-1}{k}, so the product is
12⋅23⋅34⋯910.\dfrac12 \cdot \dfrac23 \cdot \dfrac34 \cdots \dfrac{9}{10}.
Every numerator cancels the previous denominator, leaving 110.\dfrac{1}{10}.
Thus, the correct answer is A .
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