Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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== Solution 3 == | == Solution 3 == | ||
| − | 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. | + | 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 and that is <math>x,y=109,33</math> which means <math>\boxed{\textbf{(D)}\ \frac{109}{33}}</math>. |
~lopkiloinm | ~lopkiloinm | ||
Revision as of 17:06, 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.
Contents
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
Solution 3
For continued fractions of form 
, the denominator y and numerator x are solutions to the Diophantine equation 
. So for this problem, the denominator y and numerator x are solutions to the Diophantine equation 
. That leaves 
answers. Since the number of 
's in the continued fraction is odd, we further narrow it down to 
which only leaves us with answer and that is 
which means 
.
~lopkiloinm
Video Solution 1 (Quick and Easy)
~Education, the Study of Everything
Video Solution 2
~Charles3829
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. 
