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
(note that this is because the equations have the same roots and the same leading coefficients.
the denominator is by vietas.
the numerator by Newton's sums OR finding and expanding the roots
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.