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

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== Solution ==  
 
== Solution ==  
Working through the parentheses with the negative powers results in <math>\textbf{(B).}</math>
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<math>2+1=3</math>, so the reciprocal is <math>\frac{1}{3}</math>. Summing <math>1</math> and repeating leads to <math>\frac{3}{4}</math>. Rinse and repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that is <math>\frac{11}{7}</math> (B)
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-Welp...
  
 
== See Also ==
 
== See Also ==

Revision as of 21:01, 8 February 2018

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

$2+1=3$, so the reciprocal is $\frac{1}{3}$. Summing $1$ and repeating leads to $\frac{3}{4}$. Rinse and repeat to get $\frac{4}{7}$. Adding $1$ to that is $\frac{11}{7}$ (B)

-Welp...

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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