Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
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Let <math>c=\frac{2\pi}{p}</math> where <math>p</math> is an odd prime number and <math>n</math> is a relatively prime number to <math>p</math>. | Let <math>c=\frac{2\pi}{p}</math> where <math>p</math> is an odd prime number and <math>n</math> is a relatively prime number to <math>p</math>. | ||
− | Then <math>\dfrac{\sin(nc)\sin(2nc)\ldots\sin(\frac{p-1}{2}nc)}{\sin(c)\sin(2c)\ldots\sin(\frac{p-1}{2}c)}</math> is the Legendre symbol of <math>n</math> modulo <math>p</math> as famous German Number Theoretician Ferdinand Gotthold Max Eisenstein used to prove the Legendre symbol. | + | Then <math>\dfrac{\sin(nc)\sin(2nc)\ldots\sin(\frac{p-1}{2}nc)}{\sin(c)\sin(2c)\ldots\sin(\frac{p-1}{2}c)}</math> is the Legendre symbol of <math>n</math> modulo <math>p</math> as famous German Number Theoretician Ferdinand Gotthold Max Eisenstein used to prove the Legendre symbol. Any sensible person who researched Legendre symbols on wikipedia right before the competition would have gotten it in 5 seconds. |
https://en.wikipedia.org/wiki/Legendre_symbol#cite_ref-7 | https://en.wikipedia.org/wiki/Legendre_symbol#cite_ref-7 | ||
Revision as of 22:22, 16 November 2022
Contents
Problem
Let What is the value of
Solution
Plugging in , we get
Since
and
we get
~kingofpineapplz ~Ziyao7294 (minor edit)
Solution 2
Let where
is an odd prime number and
is a relatively prime number to
.
Then is the Legendre symbol of
modulo
as famous German Number Theoretician Ferdinand Gotthold Max Eisenstein used to prove the Legendre symbol. Any sensible person who researched Legendre symbols on wikipedia right before the competition would have gotten it in 5 seconds.
https://en.wikipedia.org/wiki/Legendre_symbol#cite_ref-7
~Lopkiloinm
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.