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

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(Note: integer solutions increase exponentially so our next solution will have a numerator greater than <math>3^2(109)</math> so when you don't see numerators greater than <math>3^2(109)</math> in the answer choices, this method should be fine)
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(Note: Integer solutions increase exponentially, so our next solution will have a numerator greater than <math>3^2(109)</math>. Therefore, when you don't see numerators greater than <math>3^2(109)</math> in the answer choices, this method should be fine.)
  
 
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==Video Solution 4==
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https://youtu.be/0b8OGBp1Ew0
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==Video Solution 5==
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https://www.youtube.com/watch?v=PgJcNkO8Fh8
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[[Category: Introductory Algebra Problems]]

Latest revision as of 20:30, 19 October 2025

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 linear combinations of $x, y$ that solve Pell's Equation. Specifically, 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 $3^2(109)$. Therefore, 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

Video Solution 4

https://youtu.be/0b8OGBp1Ew0

Video Solution 5

https://www.youtube.com/watch?v=PgJcNkO8Fh8

~Math4All999

See Also

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. AMC Logo.png

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