2024 AIME II Problems/Problem 13
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
whenever
is a polynomial of real coefficients.) Next, notice that
which means
. So
And we are done. Alternatively, to add some geometric flavor, we can also compute
by law of cosines.
-- 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 |
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