Difference between revisions of "2019 AMC 12A Problems/Problem 21"
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~ Baolan | ~ Baolan | ||
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+ | === Solution 4 (this is what people would write down on their scratch paper) === | ||
+ | <math>z=\mathrm{cis }(\pi/4)</math> | ||
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+ | Perfect squares mod 8: <math>1,4,1,0,1,4,1,0,1,4,1,0</math> | ||
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+ | <math>1/z=\overline{z}=\mathrm{cis }(7\pi/4)</math> | ||
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+ | <math>6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}</math> | ||
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+ | ~ MathIsFun286 | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Revision as of 23:12, 17 June 2021
Contents
Problem
Let What is
Solutions
Solution 1
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
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 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.