AIME 2001b Problem 14 PaperAnswer There are 2n2n2n complex numbers that satisfy both z28−z8−1=0z^{28} - z^{8} - 1 = 0z28−z8−1=0 and ∣z∣=1\mid z \mid = 1∣z∣=1. These numbers have the form zm=cosθm+isinθmz_{m} = \cos\theta_{m} + i\sin\theta_{m}zm=cosθm+isinθm, where 0≤θ1<θ2<…<θ2n<3600\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 3600≤θ1<θ2<…<θ2n<360 and angles are measured in degrees. Find the value of θ2+θ4+…+θ2n\theta_{2} + \theta_{4} + \ldots + \theta_{2n}θ2+θ4+…+θ2n. Dataset text AIME 1983 to 2026. Quoted with source. Show the answer 840 Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution PreviousProblem 13 NextProblem 15 Locator aime/2001b/p14 · Tag 1DSVNNHW12 · Edit this page on GitHub · Keys: j next, k previous