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) == | ||
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~ Education, the Study of Everything | ~ Education, the Study of Everything | ||
+ | == Solution 5 == | ||
+ | We notice that <math>z = e^{\pi i/4}</math> and <math>\frac1z = \overline{z} = e^{-\pi i/4}</math>. We then see that: <cmath>z^4 = e^{\pi i} = -1.</cmath> This means that: <cmath>z^{(2n)^2} = z^{4n^2} = (-1)^{n^2}.</cmath> In the first summation, there are <math>6</math> even exponents, and the <math>-1</math>'s will cancel among those. This means that: <cmath>\sum_{k=1}^{12} z^{k^2} = \sum_{m=0}^5 z^{(2m+1)^2}.</cmath> We can simplify <math>z^{(2m+1)^2}</math> to get: <cmath>z^{(2m+1)^2} = z^{4m^2} \cdot z^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot z.</cmath> We know that <math>m^2</math> and <math>m</math> will have the same parity so the <math>-1</math>'s multiply into a <math>1</math>, so what we get left is: <cmath>\sum_{m=0}^5 z^{(2m+1)^2} = \sum_{m=0}^5 z = 6z.</cmath> Now, for the conjugates, we notice that: <cmath>\overline{z}^4 = \overline{z^4} = -1.</cmath> This means that: <cmath>\overline{z}^{(2n)^2} = (-1)^{n^2}.</cmath> Therefore: <cmath>\sum_{k=1}^{12}\overline{z}^{k^2} = \sum_{m=0}^5 \overline{z}^{(2m+1)^2}.</cmath> Now, we see that: <cmath>\overline{z}^{(2m+1)^2} = \overline{z}^{4m^2} \cdot \overline{z}^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot \overline{z}.</cmath> Again, <math>m^2</math> and <math>m</math> have the same parity, so the <math>-1</math>'s multiply into a <math>1</math>, leaving us with <math>\overline{z}</math>. Therefore: <cmath>\sum_{m=0}^5 \overline{z}^{(2m+1)^2} = \sum_{m=0}^5 \overline{z} = 6\overline{z}.</cmath> Now, what we have is: <cmath>6z \cdot 6\overline{z} = 6e^{\pi i/4} \cdot 6e^{-\pi i/4} = 6 \cdot 6 = \boxed{\textbf{(C) }36}</cmath> | ||
+ | ~ ap246 | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Latest revision as of 11:36, 3 December 2023
Contents
[hide]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 |
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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