Re: [obm-l] IMO

["Henrique Lima Santana" <santanahenrique@xxxxxxxxxxx>]

a hp eh a seguinte,
www.kalva.demon.co.uk
falou
henrique





>From: "amurpe" <amurpe@bol.com.br>
>Reply-To: obm-l@mat.puc-rio.br
>To: obm-l@mat.puc-rio.br
>Subject: Re: [obm-l] IMO
>Date: Sun,  9 Feb 2003 08:37:58 -0200
>
> > 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: obm-l@mat.puc-rio.br
> >   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.
> >
> >
> >
> >
> >
> >
> >
> > -------------------------------------------------------
>-----------------------
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> >   O serviço de busca mais completo da Internet. O que v
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> >
>
>
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