Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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<math>\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}</math> | <math>\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}</math> | ||
| − | == Solution == | + | == Solution 1 == |
We have | We have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
Revision as of 04:36, 18 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 ![\[3+\frac{1}{3+\frac{1}{3+\frac13}}?\]](http://latex.artofproblemsolving.com/7/a/a/7aa51a6ff483afbbf9a1230d0817384db18841f3.png)
Solution 1
We have
~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*}](http://latex.artofproblemsolving.com/2/d/5/2d50b9330c40e3649f2dc73de9af7b1dade3b1db.png)
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*}](http://latex.artofproblemsolving.com/8/f/6/8f6bc023431782d6b8aff8f4aaf7a5e01e5ec9a7.png)
~lopkiloinm
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
