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Difference between revisions of "2018 AMC 10A Problems/Problem 1"

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==Video Solution (Simplicity)==
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https://www.youtube.com/watch?v=g0o4BbECwlk&t=16s
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==Video Solution (HOW TO THINK CREATIVELY!)==
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https://youtu.be/19mpsCcQzY0
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~Education, the Study of Everything
  
 
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https://youtu.be/19mpsCcQzY0
 
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~Education, the Study of Everything
 
  
 
== See Also ==
 
== See Also ==
 
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Latest revision as of 20:46, 26 December 2023

Problem

What is the value of \[\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?\]

$\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8$

Solution

For all nonzero numbers $a,$ recall that $a^{-1}=\frac1a$ is the reciprocal of $a.$

The original expression becomes \begin{align*} \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 &= \left(\left(3^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac13+1\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\left(\frac43\right)^{-1}+1\right)^{-1}+1 \\ &= \left(\frac34+1\right)^{-1}+1 \\ &= \left(\frac74\right)^{-1}+1 \\ &= \frac47+1 \\ &= \boxed{\textbf{(B) }\frac{11}7}. \end{align*} ~MRENTHUSIASM

Video Solution (Simplicity)

https://www.youtube.com/watch?v=g0o4BbECwlk&t=16s

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/19mpsCcQzY0

~Education, the Study of Everything

Video Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/cat3yTIpX4k

~savannahsolver

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
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First Problem 		Followed by
Problem 2
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All AMC 10 Problems and Solutions

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