Difference between revisions of "2024 AIME II Problems/Problem 13"
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==Solution 2== | ==Solution 2== | ||
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so the answer is <math>\boxed{321}</math> | so the answer is <math>\boxed{321}</math> | ||
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==Solution 3== | ==Solution 3== | ||
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\end{align*} | \end{align*} | ||
− | Therefore, the answer is \boxed{\textbf{(321) }}. | + | Therefore, the answer is <math>\boxed{\textbf{(321) }}</math>. |
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | ==Solution 4== | ||
+ | |||
+ | |||
+ | Since <math>\omega \ne 1</math> is a <math>13^\text{th}</math> root of unity, and <math>13</math> is a prime, we have | ||
+ | <cmath> | ||
+ | f(z) \coloneqq z^{13} - 1 = \prod_{k = 0}^{12}(z - \omega^k) | ||
+ | </cmath> | ||
+ | by the Fundamental Theorem of Algebra. Next, observe that the quadratic <math>2 - 2z + z^2</math> factors as | ||
+ | <cmath> | ||
+ | 2 - 2z + z^2 = (1 - i - z)(1 + i - z). | ||
+ | </cmath> | ||
+ | Take the product of the above identity over <math>z \in \{1, \omega, \omega^2, \dots, \omega^{12} \}</math> to get the product of interest \begin{align*} | ||
+ | P &:= \prod_{k = 0}^{12}(2 - 2\omega^k + \omega^{2k}) \ | ||
+ | &= \prod_{k = 0}^{12}(1 - i - \omega^k) \cdot \prod_{k = 0}^{12}(1 + i - \omega^k) \ | ||
+ | &= f(1-i) \cdot f(1+i) \ | ||
+ | &= \overline{f(1+i)} \cdot f(1+i) \ | ||
+ | P &= \big| f(1+i) \big|^2. | ||
+ | \end{align*} | ||
+ | (Here, we use the fact that <math>f(\overline{z}) = \overline{f(z)}</math> whenever <math>f(z)</math> is a polynomial of real coefficients.) Next, notice that | ||
+ | <cmath> | ||
+ | (1+i)^{13} = (1+i)(1+i)^{12} = (1+i)\big( (1+i)^2 \big)^6 = (1+i)(2i)^6 = -64 - 64i | ||
+ | </cmath> | ||
+ | which means <math>f(1+i) = (1+i)^{13} - 1 = -65 - 64i</math>. So | ||
+ | <cmath> | ||
+ | P = \big| f(1+i) \big|^2 = \big| -65 - 64i \big|^2 = 65^2 + 64^2 = 8321 \equiv \boxed{321} \pmod{1000}. | ||
+ | </cmath> | ||
+ | And we are done. Alternatively, to add some geometric flavor, we can also compute <math>\big| f(1+i) \big|^2 = \big| (1+i)^{13} - 1 \big|^2</math> by law of cosines. | ||
+ | |||
+ | -- VensL. | ||
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+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 11:55, 6 September 2024
Contents
[hide]Problem
Let be a 13th root of unity. Find the remainder when is divided by 1000.
Solution 1
Now, we consider the polynomial whose roots are the 13th roots of unity. Taking our rewritten product from to , we see that both instances of cycle through each of the 13th roots. Then, our answer is:
~Mqnic_
Solution 2
To find , where and , rewrite this is as
where and are the roots of the quadratic .
Grouping the 's and 's results in
the denomiator by vietas.
the numerator by newtons sums
so the answer is
~resources
Solution 3
Denote for .
Thus, for , is a permutation of .
We have
Note that are all zeros of the polynomial . Thus,
Plugging this into Equation (1), we get
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4
Since is a root of unity, and is a prime, we have
by the Fundamental Theorem of Algebra. Next, observe that the quadratic factors as
Take the product of the above identity over to get the product of interest
-- VensL.
Video Solution
https://youtu.be/aSD8Xz0dAI8?si=PUDeOrRg-0bVXNpp
~MathProblemSolvingSkills.com
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.