Difference between revisions of "2009 AMC 12A Problems/Problem 2"

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This is choice <math>\boxed{\text{C}}</math>.
 
This is choice <math>\boxed{\text{C}}</math>.
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Interesting sidenote: The continued fraction <math>1 + \frac {1}{1 + \frac {1}{1 + 1....}}</math> is equal to the golden ratio, or <math>\frac{1+\sqrt{5}}{2}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 09:19, 8 June 2021

The following problem is from both the 2009 AMC 12A #2 and 2009 AMC 10A #3, so both problems redirect to this page.

Problem

Which of the following is equal to $1 + \frac {1}{1 + \frac {1}{1 + 1}}$?

$\textbf{(A)}\ \frac {5}{4} \qquad \textbf{(B)}\ \frac {3}{2} \qquad \textbf{(C)}\ \frac {5}{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution

We compute:

\begin{align*} 1 + \frac {1}{1 + \frac {1}{1 + 1}} &= 1 + \frac {1}{1 + \frac {1}{1 + 1}} \\ &= 1 + \frac {1}{1 + \frac 12} \\ &= 1 + \frac {1}{\frac 32} \\ &= 1 + \frac 23 \\ &= \frac 53 \end{align*}

This is choice $\boxed{\text{C}}$.

Interesting sidenote: The continued fraction $1 + \frac {1}{1 + \frac {1}{1 + 1....}}$ is equal to the golden ratio, or $\frac{1+\sqrt{5}}{2}$.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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
2009 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 12 Problems and Solutions

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