Difference between revisions of "2021 Fall AMC 12B Problems/Problem 13"
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==Solution 3== | ==Solution 3== | ||
− | We have that <math>5^2 \equiv 3 \pmod{11}</math>, so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we is the answer from Solution 2 is <math>\boxed{\textbf{(E)}\ 1}</math> | + | We have that <math>5^2 \equiv 3 \pmod{11}</math>, so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we know is the answer from Solution 2 is <math>\boxed{\textbf{(E)}\ 1}</math> |
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+ | ==Solution 4== | ||
+ | We have <math>\zeta</math> be an 11th primitive root of unity. Then the quotient becomes <cmath>\frac{(\zeta^3-\zeta^{-3})(\zeta^6-\zeta^{-6})(\zeta^9-\zeta^{-9})(\zeta^{12}-\zeta^{-12})(\zeta^{15}-\zeta^{-15})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}</cmath> which we can use modular arithmetic to become <cmath>\frac{(\zeta^3-\zeta^{-3})(\zeta^{-5}-\zeta^{5})(\zeta^{-2}-\zeta^{2})(\zeta^{1}-\zeta^{-1})(\zeta^{4}-\zeta^{-4})}{(\zeta^1-\zeta^{-1})(\zeta^2-\zeta^{-2})(\zeta^3-\zeta^{-3})(\zeta^{4}-\zeta^{-4})(\zeta^{5}-\zeta^{-5})}</cmath> and we see that is <math>\boxed{\textbf{(E)}\ 1}</math> | ||
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+ | ~Lopkiloinm | ||
==Video Solution (Just 2 min!)== | ==Video Solution (Just 2 min!)== |
Latest revision as of 03:05, 1 February 2025
Contents
Problem
Let What is the value of
Solution
Plugging in , we get
Since
and
we get
~kingofpineapplz ~Ziyao7294 (minor edit)
Solution 2
Eisenstein used such a quotient in his proof of quadratic reciprocity. Let where
is an odd prime number and
is any integer.
Then is the Legendre symbol
. Legendre symbol is calculated using quadratic reciprocity which is
. The Legendre symbol
~Lopkiloinm
Solution 3
We have that , so 3 is a quadratic residue mod 11. For quadratic residues, their Legendre symbol which we know is the answer from Solution 2 is
Solution 4
We have be an 11th primitive root of unity. Then the quotient becomes
which we can use modular arithmetic to become
and we see that is
~Lopkiloinm
Video Solution (Just 2 min!)
~Education, the Study of Everything
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.