Difference between revisions of "2019 AMC 12A Problems/Problem 21"
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== Solution 3 (Bashing) == | == Solution 3 (Bashing) == | ||
We first calculate that <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they cancel out add up to zero. Now we can simplify the expression to <math>\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)</math>. Then, we calculate the first few odd powers of <math>z</math>. We notice that <math>z^1 = z^9</math>, so the values cycle after every 8th power. Since all of the odd squares are a multiple of <math>8</math> away from each other, <math>z^1 = z^9 = z^{25} = ... = z^{121}</math>, so <math>z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}</math>, and <math>\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}</math>. When multiplied together, we get <math>6 \cdot 6 = \boxed{\textbf{(C) } 36}</math> as our answer. | We first calculate that <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they cancel out add up to zero. Now we can simplify the expression to <math>\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)</math>. Then, we calculate the first few odd powers of <math>z</math>. We notice that <math>z^1 = z^9</math>, so the values cycle after every 8th power. Since all of the odd squares are a multiple of <math>8</math> away from each other, <math>z^1 = z^9 = z^{25} = ... = z^{121}</math>, so <math>z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}</math>, and <math>\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}</math>. When multiplied together, we get <math>6 \cdot 6 = \boxed{\textbf{(C) } 36}</math> as our answer. | ||
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== Solution 4 (this is what people would write down on their scratch paper) == | == Solution 4 (this is what people would write down on their scratch paper) == |
Revision as of 03:20, 23 May 2023
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
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.