Difference between revisions of "2022 AMC 10A Problems/Problem 1"

(Solution 2)
(Solution 2)
Line 22: Line 22:
 
where
 
where
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
[q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]
+
[q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\
[3]&=3
+
[3]&=3\\
[3,3]&=3(3)+1=10
+
[3,3]&=3(3)+1=10\\
[3,3,3]&=3(10)+3=33
+
[3,3,3]&=3(10)+3=33\\
 
[3,3,3,3]&=3(33)+10=109
 
[3,3,3,3]&=3(33)+10=109
 
\end{align*}</cmath>
 
\end{align*}</cmath>

Revision as of 00:05, 17 November 2022

The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page.

Problem

What is the value of \[3+\frac{1}{3+\frac{1}{3+\frac13}}?\] $\textbf{(A)}\ \frac{31}{10}\qquad\textbf{(B)}\ \frac{49}{15}\qquad\textbf{(C)}\ \frac{33}{10}\qquad\textbf{(D)}\ \frac{109}{33}\qquad\textbf{(E)}\ \frac{15}{4}$

Solution

We have \begin{align*} 3+\frac{1}{3+\frac{1}{3+\frac13}} &= 3+\frac{1}{3+\frac{1}{\left(\frac{10}{3}\right)}} \\ &= 3+\frac{1}{3+\frac{3}{10}} \\ &= 3+\frac{1}{\left(\frac{33}{10}\right)} \\ &= 3+\frac{10}{33} \\ &= \boxed{\textbf{(D)}\ \frac{109}{33}}. \end{align*} ~MRENTHUSIASM

Solution 2

Continued fractions are expressed as \begin{align*} \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} \end{align*} where \begin{align*} [q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ [3]&=3\\ [3,3]&=3(3)+1=10\\ [3,3,3]&=3(10)+3=33\\ [3,3,3,3]&=3(33)+10=109 \end{align*}

Video Solution 1 (Quick and Easy)

https://youtu.be/iVvBTapX3Fs

~Education, the Study of Everything

See Also

2022 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
2022 AMC 12A (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 12 Problems and Solutions

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