MathDB

14

Part of 2013 AIME Problems

Problems(2)

Sines, Cosines, and Powers of 2... Oh My!

Source: 2013 AIME I Problem 14

3/15/2013
For π≤θ<2π\pi\leq\theta<2\pi, let
P=12cos⁡θ−14sin⁡2θ−18cos⁡3θ+116sin⁡4θ+132cos⁡5θ−164sin⁡6θ−1128cos⁡7θ+… P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots and Q=1−12sin⁡θ−14cos⁡2θ+18sin⁡3θ+116cos⁡4θ−132sin⁡5θ−164cos⁡6θ+1128sin⁡7θ+… Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots so that PQ=227\tfrac PQ = \tfrac{2\sqrt2}7. Then sin⁡θ=−mn\sin\theta = -\tfrac mn where mm and nn are relatively prime positive integers. Find m+nm+n.
trigonometryquadraticsrationumber theoryrelatively primetrig identitiesgeometric series
Sums of maximums of remainders

Source: AIME II 2013, Problem 14

4/4/2013
For positive integers nn and kk, let f(n,k)f(n,k) be the remainder when nn is divided by kk, and for n>1n>1 let F(n)=max⁡1≤k≤n2f(n,k)F(n) = \displaystyle\max_{1 \le k \le \frac{n}{2}} f(n,k). Find the remainder when ∑n=20100F(n)\displaystyle\sum_{n=20}^{100} F(n) is divided by 10001000.
inductionmodular arithmeticAMCnumber theoryAIMEpattern finding