Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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&=\boxed{\textbf{(D)}\ \frac{109}{33}} | &=\boxed{\textbf{(D)}\ \frac{109}{33}} | ||
\end{align*}</cmath>~lopkiloinm | \end{align*}</cmath>~lopkiloinm | ||
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== Solution 3 == | == Solution 3 == |
Revision as of 10:02, 17 May 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 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.