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.
~ Baolan
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
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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