2019 AMC 12A Problems/Problem 21
Contents
Problem
Let What is
Solutions 1(Using Modular Functions)
Note that .
Also note that for all positive integers because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo .
and are all
and are all
and are all
and are all
Therefore,
The term thus simplifies to , while the term simplifies to . Upon multiplication, the cancels out and leaves us with .
Solution 2(Using Magnitudes and Conjugates to our Advantage)
It is well known that if then . Therefore, we have that the desired expression is equal to We know that so . Then, by De Moivre's Theorem, we have which can easily be computed as .
Solution 3 (Bashing)
We first calculate that . After a bit of calculation for the other even powers of , we realize that they cancel out add up to zero. Now we can simplify the expression to . Then, we calculate the first few odd powers of . We notice that , so the values cycle after every 8th power. Since all of the odd squares are a multiple of away from each other, , so , and . When multiplied together, we get as our answer.
Solution 4 (this is what people would write down on their scratch paper)
Perfect squares mod 8:
~ MathIsFun286
Video Solution1
~ Education, the Study of Everything
Solution 5
We notice that and . We then see that: This means that: In the first summation, there are even exponents, and the 's will cancel among those. This means that: We can simplify to get: We know that and will have the same parity so the 's multiply into a , so what we get left is: Now, for the conjugates, we notice that: This means that: Therefore: Now, we see that: Again, and have the same parity, so the 's multiply into a , leaving us with . Therefore: Now, what we have is:
~ ap246
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2019amc12a/493
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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