2005 AIME II Problem 9

Below is the professionally curated solution for Problem 9 of the 2005 AIME II, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AIME II solutions, or check the answer key.

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Concepts:De Moivre’s Theoremcomplex numbertrigonometric identity

Difficulty rating: 2460

9.

For how many positive integers nn less than or equal to 10001000 is (sint+icost)n=sinnt+icosnt(\sin t + i \cos t)^n = \sin nt + i \cos nt true for all real t?t?

Solution:

Since sint+icost=i(costisint)\sin t + i\cos t = i(\cos t - i \sin t) and sinnt+icosnt=i(cosntisinnt),\sin nt + i\cos nt = i(\cos nt - i\sin nt), de Moivre's theorem (applied to angle t-t) gives (sint+icost)n=in(costisint)n=in(cosntisinnt).(\sin t + i \cos t)^n = i^n(\cos t - i\sin t)^n = i^n(\cos nt - i \sin nt).

So the equation holds for all real tt exactly when in=i,i^n = i, that is, when n1(mod4).n \equiv 1 \pmod 4. The values n=1,5,9,,997n = 1, 5, 9, \ldots, 997 give exactly 250250 positive integers up to 1000.1000.

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