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

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== Solution ==
 
== Solution ==
Fr all nonzero numbers <math>a,</math> recall that <math>a^{-1}=\frac1a</math> is the reciprocal of <math>a.</math>
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For all nonzero numbers <math>a,</math> recall that <math>a^{-1}=\frac1a</math> is the reciprocal of <math>a.</math>
  
 
The original expression becomes
 
The original expression becomes

Revision as of 09:20, 6 November 2021

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 Solutions

https://youtu.be/vO-ELYmgRI8

https://youtu.be/cat3yTIpX4k

~savannahsolver

https://youtu.be/19mpsCcQzY0

~Education, the Study of Everything

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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