Difference between revisions of "2024 AIME II Problems/Problem 13"

Revision as of 08:42, 9 February 2024

Problem

Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when $\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})$ is divided by 1000.

Solution 1

$\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)$

Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$ , we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:

$((1 + i)^{13} - 1)(1 - i)^{13} - 1)$

$= (-64(1 + i) - 1)(-64(1 - i) - 1)$

$= (65 + 64i)(65 - 64i)$

$= 65^2 + 64^2$

$= 8\boxed{\textbf{321}}$

@@ Line 3: / Line 3: @@
 <cmath>\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})</cmath>
 is divided by 1000.
+==Solution 1==
+<cmath>\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)</cmath>
+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>= (-64(1 + i) - 1)(-64(1 - i) - 1)</cmath>
+<cmath>= (65 + 64i)(65 - 64i)</cmath>
+<cmath>= 65^2 + 64^2</cmath>
+<cmath>= 8\boxed{\textbf{321}} </cmath>

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Difference between revisions of "2024 AIME II Problems/Problem 13"

Revision as of 08:42, 9 February 2024

Problem

Solution 1