Difference between revisions of "2022 AMC 12B Problems/Problem 11"
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~lopkiloinm | ~lopkiloinm | ||
+ | == Solution 4 (Polynomial + Recursion) == | ||
+ | let <math>a = \frac{-1+i\sqrt{3}}{2}</math> and <math>b = \frac{-1-i\sqrt{3}}{2}</math> | ||
+ | <math>a + b = -1</math> | ||
+ | <math>a * b = 1</math> | ||
+ | Therefore a and b are the roots of <math>x^2 + x + 1 = 0</math> | ||
+ | By factor theorem <math>a^2 + a + 1 = 0</math> and <math>b^2 + b + 1 = 0</math> | ||
+ | Multiply the first equation by <math>a^{n-2}</math> and the second equation by <math>b^{n-2}</math> | ||
+ | This gives us <math>a^n + a^{n-1} + a^{n-2} = 0</math> and <math>b^n + b^{n-1} + b^{n-2} = 0</math>. | ||
+ | Adding both equations together we get <math>a^n + b^n + a^{n-1} + b^{n-1}+ a^{n-2} + b^{n-2} = 0</math> | ||
+ | This is the same as <math>f(n) + f(n-1) + f(n-2) = 0</math>. | ||
+ | Therefore <math>f(n) = -f(n-1) - f(n-2)</math> | ||
+ | Plugging in <math>n=1,2,3,4,5,6</math> we get <math>f(n) = -1, -1, 2, -1, -1, 2</math> therefore we know that if <math>n</math> is a multiple of <math>3</math>, then <math>f(n)</math> is <math>2</math>. | ||
+ | Since <math>2022</math> is a multiple of <math>3</math>, our answers is <math>E) 2.</math> | ||
+ | ~vpeddi18 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2022|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:54, 19 November 2022
Contents
Problem
Let , where
. What is
?
Solution 1
Converting both summands to exponential form,
Notice that both are scaled copies of the third roots of unity.
When we replace the summands with their exponential form, we get
When we substitute
, we get
We can rewrite
as
, how does that help?
Since any third root of unity must cube to
.
~
Solution 2 (Eisenstein Units)
The numbers and
are both
(along with
), denoted as
and
, respectively. They have the property that when they are cubed, they equal to
. Thus, we can immediately solve:
~mathboy100
Solution 3 (Third-order Homogeneous Linear Recurrence Relation)
Notice how this looks like the closed form of the Fibonacci sequence except different roots. This is motivation to turn this closed formula into a recurrence relation. The base of the exponents are the roots of the characteristic equation . So we have
Every time
is multiple of
as is true when
,
~lopkiloinm
Solution 4 (Polynomial + Recursion)
let and
Therefore a and b are the roots of
By factor theorem
and
Multiply the first equation by
and the second equation by
This gives us
and
.
Adding both equations together we get
This is the same as
.
Therefore
Plugging in
we get
therefore we know that if
is a multiple of
, then
is
.
Since
is a multiple of
, our answers is
~vpeddi18
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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.