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

Revision as of 21:28, 24 November 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 $\begin{align*} []&=1 \\ [q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ \end{align*}$

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

Solution 3

Finite continued fractions of form $n+\frac{1}{n+\frac{1}{n+\cdots}}=\frac{x}{y}$ have $x, y$ that solve Pell's Equation. 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 in particular, 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

(Note: integer solutions increase exponentially so our next solution will have a numerator greater than $3^2(109)$ so when you don't see numerators greater than $3^2(109)$ in the answer choices, this method should be fine)

Video Solution 1

https://youtu.be/iVvBTapX3Fs

~Education, the Study of Everything

Video Solution 2

https://youtu.be/4zoXEjrBAgk

~Charles3829

Video Solution 3

https://www.youtube.com/watch?v=7yAh4MtJ8a8&t=222s

~Math-X

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 39: / Line 39: @@
 == Solution 3 ==
-Finite continued fractions of form <math>n+\frac{1}{n+\frac{1}{n+\cdots}}=\frac{x}{y}</math> satisfy Pell's Equation. The denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{ny}{2}\right)^2=\pm{1}</math>. So for this problem in particular, the denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}</math>. That leaves two 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{3y}{2}\right)^2=-1</math>, which only leaves us with <math>1</math> answer and that is <math>(x,y)=(109,33)</math> which means <math>\boxed{\textbf{(D)}\ \frac{109}{33}}</math>.
+Finite continued fractions of form <math>n+\frac{1}{n+\frac{1}{n+\cdots}}=\frac{x}{y}</math> have <math>x, y</math> that solve Pell's Equation. The denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{ny}{2}\right)^2=\pm{1}</math>. So for this problem in particular, the denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}</math>. That leaves two 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{3y}{2}\right)^2=-1</math>, which only leaves us with <math>1</math> answer and that is <math>(x,y)=(109,33)</math> which means <math>\boxed{\textbf{(D)}\ \frac{109}{33}}</math>.
 ~lopkiloinm

Page

Toolbox

Search