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Difference between revisions of "2022 AMC 12B Problems/Problem 11"

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== Solution 3 (Difference Relation) ==
 
== Solution 3 (Difference Relation) ==
 
Notice that this is a recurrence relationship where <math>a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}</math>. This recurrence relationship goes like <math>2,-1,-1,2,-1,-1,2,\ldots</math>. Every time <math>n</math> is multiple of 3 as is true when <math>n=2022</math>, <math>a_n= \boxed{\textbf{(E)} \ 2}</math>
 
Notice that this is a recurrence relationship where <math>a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}</math>. This recurrence relationship goes like <math>2,-1,-1,2,-1,-1,2,\ldots</math>. Every time <math>n</math> is multiple of 3 as is true when <math>n=2022</math>, <math>a_n= \boxed{\textbf{(E)} \ 2}</math>
 +
~lopkiloinm
  
 
==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 02:55, 19 November 2022

Problem

Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$, where $i = \sqrt{-1}$. What is $f(2022)$?

$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$

Solution 1

Converting both summands to exponential form, \[-1 + i\sqrt{3} = 2e^{\frac{2\pi i}{3}}\] \[-1 - i\sqrt{3} = 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}\]

Notice that both are scaled copies of the third roots of unity. When we replace the summands with their exponential form, we get \[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n\] When we substitute $n = 2022$, we get \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}\] We can rewrite $2022$ as $3 \cdot 674$, how does that help? \[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} =\] \[\left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} =\] \[1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}\] Since any third root of unity must cube to $1$.

~ $\color{magenta} zoomanTV$

Solution 2 (Eisenstein Units)

The numbers $\frac{-1+i\sqrt{3}}{2}$ and $\frac{-1-i\sqrt{3}}{2}$ are both $\textbf{Eisenstein Units}$ (along with $1$), denoted as $\omega$ and $\omega^2$, respectively. They have the property that when they are cubed, they equal to $1$. Thus, we can immediately solve:

\[\omega^{2022} + \omega^{2 \cdot 2022}\] \[= \omega^{3 * 674} + \omega^{3 \cdot 2 \cdot 674}\] \[= 1^{674} + 1^{2 \cdot 674}\] \[= \boxed{\textbf{(E)} \ 2}\]

~mathboy100

Solution 3 (Difference Relation)

Notice that this is a recurrence relationship where $a_0=2,a_1=-1,a_n=-a_{n-1}-a_{n-2}$. This recurrence relationship goes like $2,-1,-1,2,-1,-1,2,\ldots$. Every time $n$ is multiple of 3 as is true when $n=2022$, $a_n= \boxed{\textbf{(E)} \ 2}$ ~lopkiloinm

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

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