Difference between revisions of "2022 AMC 10A Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
− | Continued fractions | + | Continued fractions with integer parts <math>q_i</math> and numerators all <math>1</math> can be calculated as |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} | \dfrac{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]} | ||
Line 23: | Line 23: | ||
where | where | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
+ | []&=1 \\ | ||
[q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ | [q_0,q_1,q_2,\ldots,q_n]&=q_0[q_1,q_2,\ldots,q_n]+[q_2,\ldots,q_n]\\ | ||
− | [3]&=3\\ | + | \end{align*}</cmath> |
+ | |||
+ | <cmath>\begin{align*} | ||
+ | [3]&=3(1) = 3\\ | ||
[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\\ | ||
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\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{[q_0,q_1,q_2,\ldots,q_n]}{[q_1,q_2,\ldots,q_n]}&=\dfrac{[3,3,3,3]}{[3,3,3]}\\ | ||
&=\boxed{\textbf{(D)}\ \frac{109}{33}} | &=\boxed{\textbf{(D)}\ \frac{109}{33}} | ||
− | \end{align*}</cmath>~lopkiloinm | + | \end{align*}</cmath> |
+ | ~lopkiloinm | ||
== Solution 3 == | == Solution 3 == | ||
− | + | ||
+ | Finite continued fractions of form <math>n+\frac{1}{n+\frac{1}{n+\cdots}}=\frac{x}{y}</math> have linear combinations of <math>x, y</math> that solve Pell's Equation. Specifically, the denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>(n^2+4)\left(\frac{y}{2}\right)^2-\left(x-\frac{ny}{2}\right)^2=\pm{1}</math>. So for this problem in particular, the denominator <math>y</math> and numerator <math>x</math> are solutions to the Diophantine equation <math>13\left(\frac{y}{2}\right)^2-\left(x-\frac{3y}{2}\right)^2=\pm{1}</math>. That leaves two 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{3y}{2}\right)^2=-1</math>, which only leaves us with <math>1</math> answer and that is <math>(x,y)=(109,33)</math> which means <math>\boxed{\textbf{(D)}\ \frac{109}{33}}</math>. | ||
~lopkiloinm | ~lopkiloinm | ||
− | ==Video Solution 1 | + | (Note: Integer solutions increase exponentially, so our next solution will have a numerator greater than <math>3^2(109)</math>. Therefore, when you don't see numerators greater than <math>3^2(109)</math> in the answer choices, this method should be fine.) |
+ | |||
+ | ==Video Solution 1 == | ||
https://youtu.be/iVvBTapX3Fs | https://youtu.be/iVvBTapX3Fs | ||
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~Charles3829 | ~Charles3829 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://www.youtube.com/watch?v=7yAh4MtJ8a8&t=222s | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution 4== | ||
+ | https://youtu.be/0b8OGBp1Ew0 | ||
+ | |||
+ | ==Video Solution 5== | ||
+ | https://www.youtube.com/watch?v=PgJcNkO8Fh8 | ||
+ | |||
+ | ~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.