Difference between revisions of "2022 AMC 10A Problems/Problem 1"
Lopkiloinm (talk | contribs) (→Solution 2) |
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[3,3]&=3(3)+1=10\\ | [3,3]&=3(3)+1=10\\ | ||
[3,3,3]&=3(10)+3=33\\ | [3,3,3]&=3(10)+3=33\\ | ||
− | [3,3,3,3]&=3(33)+10=109 | + | [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]}\\ | ||
+ | &=\dfrac{109}{33} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 01:06, 17 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
Solution
We have
~MRENTHUSIASM
Solution 2
Continued fractions are expressed as
where
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.