Difference between revisions of "2022 AMC 10A Problems/Problem 1"
Charles3829 (talk | contribs) m (→Video Solution 1 (Quick and Easy)) |
Lopkiloinm (talk | contribs) (→Solution 2) |
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&=\boxed{\textbf{(D)}\ \frac{109}{33}} | &=\boxed{\textbf{(D)}\ \frac{109}{33}} | ||
\end{align*}</cmath>~lopkiloinm | \end{align*}</cmath>~lopkiloinm | ||
+ | |||
+ | == Solution 2 == | ||
+ | The denominator y and numerator x must be such that $\frac{13y}{2}^2-(x-\frac{y}{2})^2=\pm{1} | ||
==Video Solution 1 (Quick and Easy)== | ==Video Solution 1 (Quick and Easy)== |
Revision as of 17:44, 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
Solution 1
We have
~MRENTHUSIASM
Solution 2
Continued fractions are expressed as
where
~lopkiloinm
Solution 2
The denominator y and numerator x must be such that $\frac{13y}{2}^2-(x-\frac{y}{2})^2=\pm{1}
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.