> Acho que nao tem muito a ver voce ficar inundando a lis ta com problemas resolvidos.. A maioria das pessoas aqui conhece o site do Kalva, e lá há diversos problemas resol vidos, de diversos níveis de dificuldade.. > ----- Original Message ----- > From: Johann Peter Gustav Lejeune Dirichlet > To: [EMAIL PROTECTED] > Sent: Thursday, February 06, 2003 3:26 PM > Subject: [obm-l] IMO > > > Problem 3 > > The set of all positive integers is the union of two disjoint subsets {f(1), f(2), f(3), ... }, {g(1), g(2), g (3), ... }, where f(1) < f(2) < f(3) < ..., and g(1) < g (2) < g(3) < ... , and g(n) = f(f (n)) + 1 for n = 1, 2, 3, ... . Determine f(240). > > > Solution > Alô pessoal , gosraia de saber o site do KAlva .por favor me enviem .um abraço.Amurpe > > Let F = {f(1), f(2), f(3), ... }, G = {g(1), g(2), g (3), ... }, Nn = {1, 2, 3, ... , n}. f(1) >= 1, so f(f (1)) >= 1 and hence g (1) >= 2. So 1 is not in G, and hence must be in F. It mu st be the smallest element of F and so f(1) = 1. Hence g (1) = 2. We can never have two successive integers n and n+1 in G, because if g(m) = n+1, then f (something) = n and so n is in F and G. Contradiction. In particular, 3 must be in F, and so f(2) = 3. > > Suppose f(n) = k. Then g(n) = f(k) + 1. So |Nf(k) +1 Ç G| = n. But |Nf(k)+1 Ç F| = k, so n + k = f (k) + 1, or f(k) = n + k - 1. Hence g (n) = n + k. So n + k + 1 must be in F and hence f (k+1) = n + k + 1. This so given the value of f for n we can find it for k and k+1. > > Using k+1 each time, we get, successively, f (2) = 3, f(4) = 6, f(7) = 11, f(12) = 19, f(20) = 32, f (33) = 53, f(54) = 87, f(88) = 142, f(143) = 231, f (232) = 375, which is not much help. Trying again with k, we get: f(3) = 4, f(4) = 6, f(6) = 9, f(9) = 14, f (14) = 22, f(22) = 35, f(35) = 56, f(56) = 90, f (90) = 145, f (145) = 234. Still not right, but we can try backing up s lightly and using k+1: f (146) = 236. Still not right, we need to back up further: f(91) = 147, f(148) = 239, f(240) = 388. > > > > > > > > ------------------------------------------------------- ----------------------- > Busca Yahoo! > O serviço de busca mais completo da Internet. O que v ocê pensar o Yahoo! encontra. >
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