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

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== Problem ==
 
== Problem ==
 
What is the value of
 
What is the value of
<cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath><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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<cmath>\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?</cmath>
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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>
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== Solution ==  
 
== Solution ==  
Working through the parentheses with the negative powers results in <math>\textbf{(B).}</math>
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<cmath> \left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 </cmath>
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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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<cmath> =\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1 </cmath>
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<cmath> =\left(\frac{3}{4}+1\right)^{-1}+1 </cmath>
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<cmath> =\left(\frac{7}{4}\right)^{-1}+1 </cmath>
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<cmath> =\frac{4}{7}+1 </cmath>
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<cmath> =\frac{11}{7} </cmath>
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Therefore, the answer is <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>.
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== Video Solutions ==
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https://youtu.be/vO-ELYmgRI8
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https://youtu.be/cat3yTIpX4k
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~savannahsolver
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https://youtu.be/19mpsCcQzY0
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~Education, the Study of Everything
  
 
== See Also ==
 
== See Also ==
 
 
{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2018|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:11, 29 November 2020

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

\[\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1\] \[=\left(\left(3)^{-1}+1\right)^{-1}+1\right)^{-1}+1\] \[=\left(\left(\frac{1}{3}+1\right)^{-1}+1\right)^{-1}+1\] \[=\left(\left(\frac{4}{3}\right)^{-1}+1\right)^{-1}+1\] \[=\left(\frac{3}{4}+1\right)^{-1}+1\] \[=\left(\frac{7}{4}\right)^{-1}+1\] \[=\frac{4}{7}+1\] \[=\frac{11}{7}\] Therefore, the answer is $\boxed{\textbf{(B) } \frac{11}{7} }$.

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
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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