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Difference between revisions of "2018 AMC 8 Problems/Problem 2"
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Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus, the answer would be <math>\boxed{\textbf{(D) }7}</math>.
 
Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus, the answer would be <math>\boxed{\textbf{(D) }7}</math>.
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Also, you can also make the fractions <math>\frac{7!}{6!}</math>, which also yields <math>\boxed{\textbf{(D)}7}</math>
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== Video Solution (CRITICAL THINKING!!!)==
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https://youtu.be/OFsrMjvR950
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==
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{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

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


See also

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

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