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

m (Solution 2)
m (Solution 2)
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[]&=1
 
[]&=1
 
[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]\\
 +
end{align*}</cmath>
  
 +
<cmath>\begin{align*}
 
[3]&=3\\
 
[3]&=3\\
 
[3,3]&=3(3)+1=10\\
 
[3,3]&=3(3)+1=10\\
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\dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\
 
\dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\
 
&=\boxed{\textbf{(D)}\ \frac{109}{33}}
 
&=\boxed{\textbf{(D)}\ \frac{109}{33}}
\end{align*}</cmath>~lopkiloinm
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\end{align*}</cmath>
 +
~lopkiloinm
  
 
== Solution 3 ==
 
== Solution 3 ==

Revision as of 01:28, 13 August 2023

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 1

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 with integer parts $q_i$ and numerators all $1$ can be calculated 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*}
[]&=1
[q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\
end{align*} (Error compiling LaTeX. Unknown error_msg)

\begin{align*} [3]&=3\\ [3,3]&=3(3)+1=10\\ [3,3,3]&=3(10)+3=33\\ [3,3,3,3]&=3(33)+10=109\\ \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\ &=\boxed{\textbf{(D)}\ \frac{109}{33}} \end{align*} ~lopkiloinm

Solution 3

It is well known that for continued fractions of form $n+\frac{1}{n+\cdots}$, the denominator $y$ and numerator $x$ are solutions to the Diophantine equation $(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{ny}{2}\right)^2=\pm{1}$. So for this problem, the denominator $y$ and numerator $x$ are solutions to the Diophantine equation $13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}$. That leaves two answers. Since the number of $1$'s in the continued fraction is odd, we further narrow it down to $13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=-1$, which only leaves us with $1$ answer and that is $(x,y)=(109,33)$ which means $\boxed{\textbf{(D)}\ \frac{109}{33}}$.

~lopkiloinm

Video Solution 1 (Quick and Easy)

https://youtu.be/iVvBTapX3Fs

~Education, the Study of Everything

Video Solution 2

https://youtu.be/4zoXEjrBAgk

~Charles3829

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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