Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
− | Continued fractions | + | Continued fractions with integer parts <math>q_i</math> and numerators all <math>1</math> can be calculated as |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} | \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} | ||
Line 23: | Line 23: | ||
where | where | ||
<cmath>\begin{align*} | <cmath>\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]\\ | [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]&=3(3)+1=10\\ | [3,3]&=3(3)+1=10\\ |
Revision as of 01:27, 13 August 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
Solution 1
We have ~MRENTHUSIASM
Solution 2
Continued fractions with integer parts and numerators all can be calculated as 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]\\ [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*} (Error compiling LaTeX. Unknown error_msg)
~lopkiloinm
Solution 3
It is well known that for continued fractions of form , the denominator and numerator are solutions to the Diophantine equation . So for this problem, the denominator and numerator are solutions to the Diophantine equation . That leaves two 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.