2022 AMC 10A Problems/Problem 1: Difference between revisions
| Line 23: | Line 23: | ||
where | where | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
[]&=1 | []&=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]\\ | ||
end{align*}</cmath> | end{align*}</cmath> | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
[3]&=3\\ | [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\\ | ||
Revision as of 01:29, 13 August 2023
- The following problem is from both the 2022 AMC 10A #1 and 2022 AMC 12A #1, so both problems redirect to this page.
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
\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]\\
end{align*} (Error compiling LaTeX. Unknown error_msg)
~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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing