Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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== Solution 3 == | == Solution 3 == | ||
− | The denominator y and numerator x | + | The denominator y and numerator x are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=\pm{1}</math>. That leaves <math>2</math> answers. Since the number of <math>1</math>'s in the continued fraction is odd, we further narrow it down to <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{y}{2}\right)^2=-1</math> which only leaves us with <math>1</math> answer. |
~lopkiloinm | ~lopkiloinm |
Revision as of 17:55, 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 3
The denominator y and numerator x are solutions to the Diophantine equation . That leaves
answers. Since the number of
's in the continued fraction is odd, we further narrow it down to
which only leaves us with
answer.
~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.