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

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== Solution ==  
 
== Solution ==  
<math>2+1=3</math>, and the reciprocal is <math>\frac{1}{3}</math>. Adding <math>1</math> gives <math>\frac{4}{3}</math> and the reciprocal is <math>\frac{3}{4}</math>. Repeat to get <math>\frac{4}{7}</math>. Adding <math>1</math> to that gives <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>
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We will start with <math>2+1=3</math> and then apply the operation "invert and add one" three times. These iterations yield (after <math>3</math>): <math>\frac{4}{3}</math>, <math>\frac{7}{4}</math>, and finally <math>\boxed{\textbf{(B) } \frac{11}{7} }</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 03:11, 10 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

We will start with $2+1=3$ and then apply the operation "invert and add one" three times. These iterations yield (after $3$): $\frac{4}{3}$, $\frac{7}{4}$, and finally $\boxed{\textbf{(B) } \frac{11}{7} }$

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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