2022 AMC 10A Problems/Problem 1: Difference between revisions

Revision as of 17:03, 12 January 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 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 $\begin{align*} [q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ [3]&=3\\ [3,3]&=3(3)+1=10\\ [3,3,3]&=3(10)+3=33\\ [3,3,3,3]&=3(33)+10=109\\ \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\ &=\boxed{\textbf{(D)}\ \frac{109}{33}} \end{align*}$ ~lopkiloinm

Solution 3

For continued fractions of form $n+\frac{1}{n+\ldots}$ , the denominator y and numerator x are solutions to the Diophantine equation $(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}$ . So for this problem, the denominator y and numerator x are solutions to the Diophantine equation $13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}$ . That leaves $2$ answers. Since the number of $1$ 's in the continued fraction is odd, we further narrow it down to $13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=-1$ which only leaves us with $1$ answer.

~lopkiloinm

Video Solution 1 (Quick and Easy)

https://youtu.be/iVvBTapX3Fs

~Education, the Study of Everything

Video Solution 2

https://youtu.be/4zoXEjrBAgk

~Charles3829

2022 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 == Solution 3 ==
-For continued fractions of form <math>x+\frac{1}{x+\ldots}</math>, the denominator y and numerator x are solutions to the Diophantine equation <math>(x^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}</math>. So for this problem, the denominator y and numerator x are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}</math>. That leaves <math>2</math> answers. Since the number of <math>1</math>'s in the continued fraction is odd, we further narrow it down to <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=-1</math> which only leaves us with <math>1</math> answer.
+For continued fractions of form <math>n+\frac{1}{n+\ldots}</math>, the denominator y and numerator x are solutions to the Diophantine equation <math>(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}</math>. So for this problem, the denominator y and numerator x are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}</math>. That leaves <math>2</math> answers. Since the number of <math>1</math>'s in the continued fraction is odd, we further narrow it down to <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=-1</math> which only leaves us with <math>1</math> answer.
 ~lopkiloinm

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