2018 AMC 8 Problem 2

Attempt Problem 2 of the 2018 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 2018 AMC 8 solutions, or check the answer key.

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

2.

What is the value of the product (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)?\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)? \end{align*}
76 \dfrac{7}{6}
43 \dfrac{4}{3}
72 \dfrac{7}{2}
7 7
8 8
Answer: D
Concepts:telescopingfraction
Small hint:
Rewrite each factor 1+1n1+\frac1n as n+1n\frac{n+1}{n}.
Big hint:
The product telescopes after the fractions are rewritten.
Video solution:
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Written solution:
Let’s first note that if we are given an expression of the form 1+1n,1 + \frac{1}{n}, we can rewrite this as nn+1n=n+1n.\frac{n}{n} + \frac{1}{n} = \frac{n+1}{n}. With that in mind, we can rewrite the expression given to us in the problem, as shown below: (1+11)⋅(1+12)⋅(1+13)⋅(1+14)⋅(1+15)⋅(1+16)=21⋅32⋅43⋅54⋅65⋅76=7\begin{align*} &\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\\ &\quad{}\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\\ &\quad{}\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)\\ &=\dfrac{2}{1} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot\dfrac{5}{4} \cdot\dfrac{6}{5} \cdot\dfrac{7}{6}\\ &=7 \end{align*} Thus, the correct answer is D.
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