2019 AMC 8 Problem 17

Attempt Problem 17 of the 2019 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 8 solutions, or check the answer key.

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

17.

What is the value of the product below?
(1⋅32⋅2)(2⋅43⋅3)(3⋅54⋅4)…(97⋅9998⋅98)(98⋅10099⋅99)\begin{align*}\displaystyle & \left(\dfrac{1\cdot 3}{2\cdot 2}\right)\left(\dfrac{2\cdot 4}{3\cdot 3}\right)\left(\dfrac{3\cdot 5}{4\cdot 4}\right) \\ & \dots\left(\dfrac{97\cdot 99}{98\cdot 98}\right)\left(\dfrac{98\cdot 100}{99\cdot 99}\right) \end{align*}
12\displaystyle \dfrac{1}{2}
5099\displaystyle \dfrac{50}{99}
98009801\displaystyle \dfrac{9800}{9801}
10099\displaystyle \dfrac{100}{99}
5050
Answer: B
Concepts:telescopingfraction
Small hint:
Rewrite each factor to line up cancellations
Big hint:
Only the first 12\frac12 and the last 10099\frac{100}{99} remain
Video solution:
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Written solution:
We can regroup all the factors as follows. 12⋅(3⋅22⋅3)(4⋅33⋅4)…(99⋅9898⋅99)⋅10099\begin{align*} & \dfrac{1}{2} \cdot \left(\dfrac{3 \cdot 2}{2\cdot 3}\right)\left(\dfrac{4\cdot 3}{3\cdot 4}\right) \\ & \dots\left(\dfrac{99\cdot 98}{98\cdot 99}\right) \cdot \dfrac{100}{99} \end{align*}
From this representation, we can see that all the middle terms cancel leaving only 12⋅10099=5099. \dfrac{1}{2} \cdot \dfrac{100}{99} = \dfrac{50}{99}.
Thus, the correct answer is B.
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