Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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==Video Solution 4== | ==Video Solution 4== | ||
https://youtu.be/0b8OGBp1Ew0 | https://youtu.be/0b8OGBp1Ew0 | ||
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+ | ==Video Solution 5== | ||
+ | https://www.youtube.com/watch?v=PgJcNkO8Fh8 | ||
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+ | ~Math4All999 | ||
== See Also == | == See Also == |
Latest revision as of 23:45, 14 September 2024
- 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
~lopkiloinm
Solution 3
Finite continued fractions of form have linear combinations of that solve Pell's Equation. Specifically, the denominator and numerator are solutions to the Diophantine equation . So for this problem in particular, 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
(Note: Integer solutions increase exponentially, so our next solution will have a numerator greater than . Therefore, when you don't see numerators greater than in the answer choices, this method should be fine.)
Video Solution 1
~Education, the Study of Everything
Video Solution 2
~Charles3829
Video Solution 3
https://www.youtube.com/watch?v=7yAh4MtJ8a8&t=222s
~Math-X
Video Solution 4
Video Solution 5
https://www.youtube.com/watch?v=PgJcNkO8Fh8
~Math4All999
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.