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[SPAM] [obm-l] Re: [obm-l] Re: [obm-l] Re: [obm-l] OBM TERCEIRA FASE â?" NÃ?VEL 3 -- 2ª questão



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Vlw douglas!

Cara, n=E3o mandei pra lugar algum, acho q eles devem ter uma =
solu=E7=E3o melhor que a minha, com certeza

Se invertermos o problema, dizendo que 2^2007 =E9 m=FAltiplo de c + x^2, =
e quisermos somente os valores de c no intervalo [0,2007], como seria?
  ----- Original Message -----=20
  From: douglas paula=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Thursday, May 29, 2008 9:14 PM
  Subject: Re: [obm-l] Re: [obm-l] Re: [obm-l] OBM TERCEIRA FASE =E2?" =
N=C3=8DVEL 3 -- 2=C2=AA quest=C3=A3o


  Vlw rodrigo muito maneira a sua solu=E7=E3o. J=E1 mandou ela pra =
eureka ?
                                                     abra=E7os

  Rodrigo Cientista <rodrigocientista@xxxxxxxxxxxx> escreveu:
    Douglas, desculpe-me, li mal o problema, a minha solu=C3=A7=C3=A3o =
segue abaixo:
    como c + x^2 =C3=A9 m=C3=BAltiplo de 2^2007, ent=C3=A3o c + x^2 =3D =
w2^2007
    partimos de duas constata=C3=A7=C3=B5es:
    a) um quadrado perfeito par =C3=A9 divis=C3=ADvel por 4
    **prova: tome x^2 par =3D=3D> x =C3=A9 par =3D=3D> x =3D 2k =3D=3D: =
x^2 =3D 4k^2
    b) um quadrado perfeito =C3=ADmpar =C3=A9 da forma 8a + 1
    **prova: tome x^2 =C3=ADmpar =3D=3D> x =C3=A9 =C3=ADmpar =3D=3D> x =
=C3=A9 da forma 2n+1 =3D=3D> x^2 =3D (2n+1)^2 =3D 4n^2 + 4n + 1 =3D =
4n(n+1) + 1, como n e n+1 s=C3=A3o consecutivos, um deles =C3=A9 par, =
logo n(n+1) por ser escrito como 2a =3D=3D> 4n(n+1) + 1 =3D 8a + 1 =3D =
x^2
    1 ) no caso em que x^2 =C3=A9 par, temos que x^2 =3D 4k^2 =3D=3D> c =
=3D w2^2007 - 4k^2, como 4 divide 2^2007 =3D=3D> 4 divide w2^2007 - 4k^2 =
=3D=3D> 4 divide c, logo c assume os valores m=C3=BAltiplos de 4 no =
intervalo [-2007, 2007] (para que sua soma com um x^2 suficientemente =
grande seja divis=C3=ADvel por 2^2007), incluindo o zero, que s=C3=A3o =
no total de 501 + 501 + 1 =3D 1003 (4 divide 2007 - 3 em 501 partes, =
mesmo racioc=C3=ADnio para 3 - 2007)
    2 ) no caso em que x^2 =C3=A9 =C3=ADmpar, temos que x^2 =3D 8a + 1 =
=3D=3D> c + 8a + 1 =3D w2^2007 =3D=3D> c + 1 =3D w2^2007 - 8a, como 8 =
divide w2^2007 - 8a =3D=3D> 8 divide c + 1, logo c assume os valores que =
somados a 1 s=C3=A3o m=C3=BAltiplos de 8 no intervalo [-2007, 2007] =
(para que sua soma com um x^2 suficientemente grande seja divis=C3=ADvel =
por 2^2007, mesmo racioc=C3=ADnio), excluindo o zero pois j=C3=A1 foi =
contado, que s=C3=A3o no total de 250 + 250 =3D 500 (8 divide 2007 - 7 =
em 250 partes, mesmo racioc=C3=ADnio para 7 - 2007)
    RESP: para 1503 inteiros c

    ----- Original Message -----=20
    From: douglas paula=20
    To: obm-l@xxxxxxxxxxxxxx=20
    Sent: Tuesday, May 27, 2008 9:44 PM
    Subject: Re: [obm-l] Re: [obm-l] OBM TERCEIRA FASE =
=C3=A2=E2,=AC=E2?o N=C3fVEL 3 -- 2=C3,=C2=AA quest=C3f=C2=A3o

    rodrigo,
    =C2 ao meu ver, c + x^2 =3D k 2^2007 , onde k =C3=A9 qq natural e k =
2^2007 n=C3=A3o =C3=A9 necessariamente igual =C3  2^n
    venho a um bom tempo quebrando a cabe=C3=A7a nessa quest=C3=A3o mas =
sem conseguir muito resultado ...
    rodrigocientista@xxxxxxxxxxxx escreveu:
    =C3=AF=C2=BB=C2=BF=20
    vou tentar,
    2^n - x^2 =3D c tal qque 1< n < 2007, como todo n=C3f=C2=BAmero pode =
ser expresso como diferen=C3f=C2=A7a de dois quadrados, s=C3f=C2=B3 =
existem "c" tal que n possa ser um quadrado, de sorte que c seja =
expresso como diferen=C3f=C2=A7a de dois quadrados

    ----- Original Message -----=20
    From: douglas paula=20
    To: obm-l@xxxxxxxxxxxxxx=20
    Sent: Saturday, May 17, 2008 11:02 PM
    Subject: [obm-l] OBM TERCEIRA FASE =C3=A2=E2,=AC=E2?o N=C3fVEL 3 -- =
2=C3,=C2=AA quest=C3f=C2=A3o

    XXIX OLIMP=C3fADA BRASILEIRA DE MATEM=C3fTICA
    TERCEIRA FASE =C3=A2=E2,=AC=E2?o N=C3fVEL 3 (Ensino M=C3f=C2=A9dio)
    PRIMEIRO DIA
    PROBLEMA 2
    Para quantos n=C3=BAmeros inteiros c, - 2007 <=3D c <=3D 2007 , =
existe um inteiro x tal que x^2 + c =C3=A9 m=C3=BAltiplo de 2^2007?=20
    algu=C3=A9m se habilita?
    grato,=20
    =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2 =C2  =
Douglas
    =
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espa=C3f=C2=A7o para armazenamento!=20

    =
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    Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite de =
espa=C3=A7o para armazenamento!
    http://br.mail.yahoo.com/

    =
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<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Vlw douglas!</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Cara, n=E3o mandei pra lugar algum, =
acho q eles devem=20
ter uma solu=E7=E3o melhor que a minha, com certeza</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Se invertermos o problema, dizendo que =
2^2007 =E9=20
m=FAltiplo de c + x^2, e quisermos somente os valores de c no intervalo =
[0,2007],=20
como seria?</FONT></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A title=3Ddouglasfogo@xxxxxxxxxxxx=20
  href=3D"mailto:douglasfogo@xxxxxxxxxxxx";>douglas paula</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
  href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Thursday, May 29, 2008 =
9:14=20
PM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: [obm-l] Re: =
[obm-l] Re:=20
  [obm-l] OBM TERCEIRA FASE =E2=80=93 N=C3=8DVEL 3 -- 2=C2=AA =
quest=C3=A3o</DIV>
  <DIV><BR></DIV>
  <DIV>Vlw&nbsp;rodrigo muito maneira&nbsp;a sua solu=E7=E3o. J=E1 =
mandou ela pra=20
  eureka ?</DIV>
  =
<DIV>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
;&nbsp;&nbsp;=20
  abra=E7os<BR><BR><B><I>Rodrigo Cientista &lt;<A=20
  =
href=3D"mailto:rodrigocientista@xxxxxxxxxxxx";>rodrigocientista@xxxxxxxxxx=
br</A>&gt;</I></B>=20
  escreveu:</DIV>
  <BLOCKQUOTE class=3Dreplbq=20
  style=3D"PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px =
solid">Douglas,=20
    desculpe-me, li mal o problema, a minha solu=C3=A7=C3=A3o segue =
abaixo:<BR>como c +=20
    x^2 =C3=A9 m=C3=BAltiplo de 2^2007, ent=C3=A3o c + x^2 =3D =
w2^2007<BR>partimos de duas=20
    constata=C3=A7=C3=B5es:<BR>a) um quadrado perfeito par =C3=A9 =
divis=C3&shy;vel por=20
    4<BR>**prova: tome x^2 par =3D=3D&gt; x =C3=A9 par =3D=3D&gt; x =3D =
2k =3D=3D: x^2 =3D=20
    4k^2<BR>b) um quadrado perfeito =C3&shy;mpar =C3=A9 da forma 8a + =
1<BR>**prova:=20
    tome x^2 =C3&shy;mpar =3D=3D&gt; x =C3=A9 =C3&shy;mpar =3D=3D&gt; x =
=C3=A9 da forma 2n+1 =3D=3D&gt;=20
    x^2 =3D (2n+1)^2 =3D 4n^2 + 4n + 1 =3D 4n(n+1) + 1, como n e n+1 =
s=C3=A3o=20
    consecutivos, um deles =C3=A9 par, logo n(n+1) por ser escrito como =
2a =3D=3D&gt;=20
    4n(n+1) + 1 =3D 8a + 1 =3D x^2<BR>1 ) no caso em que x^2 =C3=A9 par, =
temos que x^2 =3D=20
    4k^2 =3D=3D&gt; c =3D w2^2007 - 4k^2, como 4 divide 2^2007 =
=3D=3D&gt; 4 divide w2^2007=20
    - 4k^2 =3D=3D&gt; 4 divide c, logo c assume os valores =
m=C3=BAltiplos de 4 no=20
    intervalo [-2007, 2007] (para que sua soma com um x^2 =
suficientemente grande=20
    seja divis=C3&shy;vel por 2^2007), incluindo o zero, que s=C3=A3o no =
total de 501=20
    + 501 + 1 =3D 1003 (4 divide 2007 - 3 em 501 partes, mesmo =
racioc=C3&shy;nio=20
    para 3 - 2007)<BR>2 ) no caso em que x^2 =C3=A9 =C3&shy;mpar, temos =
que x^2 =3D 8a +=20
    1 =3D=3D&gt; c + 8a + 1 =3D w2^2007 =3D=3D&gt; c + 1 =3D w2^2007 - =
8a, como 8 divide=20
    w2^2007 - 8a =3D=3D&gt; 8 divide c + 1, logo c assume os valores que =
somados a 1=20
    s=C3=A3o m=C3=BAltiplos de 8 no intervalo [-2007, 2007] (para que =
sua soma com um=20
    x^2 suficientemente grande seja divis=C3&shy;vel por 2^2007, mesmo=20
    racioc=C3&shy;nio), excluindo o zero pois j=C3=A1 foi contado, que =
s=C3=A3o no total=20
    de 250 + 250 =3D 500 (8 divide 2007 - 7 em 250 partes, mesmo =
racioc=C3&shy;nio=20
    para 7 - 2007)<BR>RESP: para 1503 inteiros c<BR><BR>----- Original =
Message=20
    ----- <BR>From: douglas paula <BR>To: obm-l@xxxxxxxxxxxxxx <BR>Sent: =

    Tuesday, May 27, 2008 9:44 PM<BR>Subject: Re: [obm-l] Re: [obm-l] =
OBM=20
    TERCEIRA FASE =C3=A2=E2=82=AC=E2=80=9C N=C3=83VEL 3 -- 2=C3=82=C2=AA =

    quest=C3=83=C2=A3o<BR><BR>rodrigo,<BR>=C2&nbsp;ao meu ver, c + x^2 =
=3D k 2^2007 , onde k=20
    =C3=A9 qq natural e k 2^2007 n=C3=A3o =C3=A9 necessariamente igual =
=C3&nbsp; 2^n<BR>venho=20
    a um bom tempo quebrando a cabe=C3=A7a nessa quest=C3=A3o mas sem =
conseguir muito=20
    resultado ...<BR>rodrigocientista@xxxxxxxxxxxx =
escreveu:<BR>=C3=AF=C2=BB=C2=BF <BR>vou=20
    tentar,<BR>2^n - x^2 =3D c tal qque 1&lt; n &lt; 2007, como todo =
n=C3=83=C2=BAmero=20
    pode ser expresso como diferen=C3=83=C2=A7a de dois quadrados, =
s=C3=83=C2=B3 existem "c" tal=20
    que n possa ser um quadrado, de sorte que c seja expresso como =
diferen=C3=83=C2=A7a=20
    de dois quadrados<BR><BR>----- Original Message ----- <BR>From: =
douglas=20
    paula <BR>To: obm-l@xxxxxxxxxxxxxx <BR>Sent: Saturday, May 17, 2008 =
11:02=20
    PM<BR>Subject: [obm-l] OBM TERCEIRA FASE =C3=A2=E2=82=AC=E2=80=9C =
N=C3=83VEL 3 -- 2=C3=82=C2=AA=20
    quest=C3=83=C2=A3o<BR><BR>XXIX OLIMP=C3=83ADA BRASILEIRA DE =
MATEM=C3=83TICA<BR>TERCEIRA FASE=20
    =C3=A2=E2=82=AC=E2=80=9C N=C3=83VEL 3 (Ensino =
M=C3=83=C2=A9dio)<BR>PRIMEIRO DIA<BR>PROBLEMA 2<BR>Para=20
    quantos n=C3=BAmeros inteiros c, - 2007 &lt;=3D c &lt;=3D 2007 , =
existe um inteiro x=20
    tal que x^2 + c =C3=A9 m=C3=BAltiplo de 2^2007? <BR>algu=C3=A9m se =
habilita?<BR>grato,=20
    =
<BR>=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nb=
sp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbsp;=C2&nbs=
p;=20
    =
Douglas<BR>--------------------------------------------------------------=
------------------<BR>Abra=20
    sua conta no Yahoo! Mail, o =C3=83=C2=BAnico sem limite de =
espa=C3=83=C2=A7o para=20
    armazenamento!=20
    =
<BR><BR>-----------------------------------------------------------------=
---------------<BR>Abra=20
    sua conta no Yahoo! Mail, o =C3=BAnico sem limite de espa=C3=A7o =
para armazenamento!=20
    <BR><BR><BR>Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite =
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    =
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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
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