Difference between revisions of "2024 AIME II Problems/Problem 13"
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Now, we consider the polynomial <math>x^{13} - 1</math> whose roots are the 13th roots of unity. Taking our rewritten product from <math>0</math> to <math>12</math>, we see that both instances of <math>\omega^k</math> cycle through each of the 13th roots. Then, our answer is: | Now, we consider the polynomial <math>x^{13} - 1</math> whose roots are the 13th roots of unity. Taking our rewritten product from <math>0</math> to <math>12</math>, we see that both instances of <math>\omega^k</math> cycle through each of the 13th roots. Then, our answer is: | ||
− | <cmath>((1 + i)^{13} - 1)(1 - i)^{13} - 1)</cmath> | + | <cmath>((1 + i)^{13} - 1)((1 - i)^{13} - 1)</cmath> |
<cmath>= (-64(1 + i) - 1)(-64(1 - i) - 1)</cmath> | <cmath>= (-64(1 + i) - 1)(-64(1 - i) - 1)</cmath> | ||
Line 20: | Line 20: | ||
~Mqnic_ | ~Mqnic_ | ||
− | |||
− | |||
==Solution 2== | ==Solution 2== | ||
Line 33: | Line 31: | ||
the denomiator <math>(r-1)(s-1)=1</math> by vietas. | the denomiator <math>(r-1)(s-1)=1</math> by vietas. | ||
− | the numerator <math>(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321</math> by | + | the numerator <math>(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321</math> by Newton's sums |
so the answer is <math>\boxed{321}</math> | so the answer is <math>\boxed{321}</math> | ||
− | ~ | + | ~resources |
==Solution 3== | ==Solution 3== |
Latest revision as of 14:10, 16 January 2025
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 Newton's 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.