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

Latest revision as of 13:07, 6 June 2025

Problem

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 of the $6$ parentheses, we get:

$\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}$ .

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 $\boxed{\textbf{(D) }7}$ .

Also, you can also make the fractions $\frac{7!}{6!}$ , which also yields $\boxed{\textbf{(D)}7}$

Video Solution (CRITICAL THINKING!!!)

https://youtu.be/OFsrMjvR950

~Education, the Study of Everything

Video Solution

https://youtu.be/OeaCROQV4j4

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=3213

~ pi_is_3.14

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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		+	[[Category:Introductory Algebra Problems]]

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