Difference between revisions of "2018 AMC 8 Problems/Problem 2"

Revision as of 14:56, 27 October 2019

Problem 2

What is the value of the product $\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?$

$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Solution

By adding up the numbers in each parentheses, we have: $\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$ .

Using telescoping, most of the terms cancel out diagonally. We are left with $\frac{7}{1}$ which is equivalent to $7$ . Thus the answer would be $\boxed{\textbf{(D) }7}$ .

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Second Problem	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==See Also==
-{{AMC8 box|year=2018|before=First Problem|num-a=2}}
+{{AMC8 box|year=2018|before=Second Problem|num-a=3}}
 {{MAA Notice}}

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Difference between revisions of "2018 AMC 8 Problems/Problem 2"

Revision as of 14:56, 27 October 2019

Problem 2

Solution

See Also