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

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<math>\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 </math>
 
<math>\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 </math>
  
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== Solution ==  
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== Solution ==
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<cmath> \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath>
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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>
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<cmath> =\left(\left(3)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath>
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<cmath> =\left(\left(\frac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 </cmath>
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The original expression becomes
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<cmath> =\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1 </cmath>
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<cmath>\begin{align*}
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<cmath> =\left(\frac{3}{4}+1\right)^{-1}+1 </cmath>
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\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 &= \left(\left(3^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\
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<cmath> =\left(\frac{7}{4}\right)^{-1}+1 </cmath>
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&= \left(\left(\frac13+1\right)^{-1}+1\right)^{-1}+1 \\
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<cmath> =\frac{4}{7}+1 </cmath>
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&= \left(\left(\frac43\right)^{-1}+1\right)^{-1}+1 \\
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<cmath> =\frac{11}{7} </cmath>
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&= \left(\frac34+1\right)^{-1}+1 \\
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Therefore, the answer is <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>.
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&= \left(\frac74\right)^{-1}+1 \\
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&= \frac47+1 \\
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&= \boxed{\textbf{(B) }\frac{11}7}.
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\end{align*}</cmath>
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~MRENTHUSIASM
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==Video Solution (Simplicity)==
 +
 
 +
https://www.youtube.com/watch?v=g0o4BbECwlk&t=16s
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==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/19mpsCcQzY0
 +
 
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~Education, the Study of Everything
  
 
== Video Solutions ==
 
== Video Solutions ==

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)
Preceded by
First Problem 		Followed by
Problem 2
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 AMC 10 Problems and Solutions

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