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[SPAM] [obm-l] Re: [obm-l] Re: [obm-l] OBM TERCEIRA FASE – NÃVEL 3 -- 2ª questão
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- Subject: [SPAM] [obm-l] Re: [obm-l] Re: [obm-l] OBM TERCEIRA FASE – NÃVEL 3 -- 2ª questão
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- Date: Wed, 28 May 2008 09:58:46 -0300
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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=82=AC=E2=80=9C N=C3=83=C2=8DVEL 3 -- 2=C3=82=C2=AA =
quest=C3=83=C2=A3o
rodrigo,
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=A0 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=C3=83=C2=BAmero =
pode ser expresso como diferen=C3=83=C2=A7a de dois quadrados, =
s=C3=83=C2=B3 existem "c" tal que n possa ser um quadrado, de sorte que =
c seja expresso como diferen=C3=83=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=82=AC=E2=80=9C =
N=C3=83=C2=8DVEL 3 -- 2=C3=82=C2=AA quest=C3=83=C2=A3o
XXIX OLIMP=C3=83=C2=8DADA BRASILEIRA DE MATEM=C3=83=C2=81TICA
TERCEIRA FASE =C3=A2=E2=82=AC=E2=80=9C N=C3=83=C2=8DVEL 3 (Ensino =
M=C3=83=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
Douglas
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-
Abra sua conta no Yahoo! Mail, o =C3=83=C2=BAnico sem limite de =
espa=C3=83=C2=A7o para armazenamento!=20
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-----
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para armazenamento!
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charset="utf-8"
Content-Transfer-Encoding: quoted-printable
=EF=BB=BF<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; charset=3Dutf-8">
<META content=3D"MSHTML 6.00.2900.3314" name=3DGENERATOR></HEAD>
<BODY bgColor=3D#ffffff>
<DIV>
<DIV><FONT face=3DArial size=3D2>Douglas, desculpe-me, li mal o =
problema, a minha=20
solu=C3=A7=C3=A3o segue abaixo:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>como c + x^2 =C3=A9 m=C3=BAltiplo de =
2^2007, ent=C3=A3o c + x^2 =3D=20
w2^2007</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>partimos de duas =
constata=C3=A7=C3=B5es:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>a) um quadrado perfeito par =C3=A9 =
divis=C3=ADvel por=20
4</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>**prova: tome x^2 par =3D=3D> x =
=C3=A9 par =3D=3D> x =3D 2k=20
=3D=3D: x^2 =3D 4k^2</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>b) um quadrado perfeito =C3=ADmpar =
=C3=A9 da forma 8a +=20
1</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>**prova: tome x^2 =C3=ADmpar =3D=3D> =
x =C3=A9 =C3=ADmpar =3D=3D> x =C3=A9=20
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=20
s=C3=A3o consecutivos, um deles =C3=A9 par, logo n(n+1) por ser =
escrito como 2a=20
=3D=3D> 4n(n+1) + 1 =3D 8a + 1 =3D x^2</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>1 ) no caso em que x^2 =C3=A9 par, =
temos que x^2 =3D 4k^2=20
=3D=3D> c =3D w2^2007 - 4k^2, como 4 divide 2^2007 =3D=3D> 4 =
divide w2^2007 - 4k^2=20
=3D=3D> 4 divide c, logo c assume os valores m=C3=BAltiplos de 4 no =
intervalo [-2007,=20
2007] (para que sua soma com um x^2 suficientemente grande seja =
divis=C3=ADvel por=20
2^2007), incluindo o zero, que s=C3=A3o no total de 501 + 501 + 1 =
=3D 1003 (4=20
divide 2007 - 3 em 501 partes, mesmo racioc=C3=ADnio para 3 - =
2007)</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>2 ) no caso em que x^2 =C3=A9 =
=C3=ADmpar, temos que x^2 =3D 8a=20
+ 1 =3D=3D> c + 8a + 1 =3D w2^2007 =3D=3D> c + 1 =3D w2^2007 - 8a, =
como 8 divide=20
w2^2007 - 8a =3D=3D> 8 divide c + 1, logo c assume os valores que =
somados a 1 s=C3=A3o=20
m=C3=BAltiplos de 8 no intervalo [-2007, 2007] (para que sua soma com um =
x^2=20
suficientemente grande seja divis=C3=ADvel por 2^2007, mesmo =
racioc=C3=ADnio), excluindo o=20
zero pois j=C3=A1 foi contado, que s=C3=A3o no total de 250 + 250 =3D =
500 (8 divide 2007 - 7=20
em 250 partes, mesmo racioc=C3=ADnio para 7 - 2007)</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>RESP: para 1503 inteiros =
c</FONT></DIV></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> Tuesday, May 27, 2008 =
9:44 PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: [obm-l] Re: =
[obm-l] OBM=20
TERCEIRA FASE =C3=A2=E2=82=AC=E2=80=9C N=C3=83=C2=8DVEL 3 -- =
2=C3=82=C2=AA quest=C3=83=C2=A3o</DIV>
<DIV><BR></DIV>
<DIV>rodrigo,</DIV>
<DIV> </DIV>
<DIV> ao meu ver, c + x^2 =3D k 2^2007 , onde k =C3=A9 qq natural =
e k 2^2007 n=C3=A3o=20
=C3=A9 necessariamente igual =C3=A0 2^n</DIV>
<DIV> </DIV>
<DIV>venho a um bom tempo quebrando a cabe=C3=A7a nessa quest=C3=A3o =
mas sem conseguir=20
muito resultado ...</DIV>
<DIV><BR><B><I><A=20
=
href=3D"mailto:rodrigocientista@xxxxxxxxxxxx";>rodrigocientista@xxxxxxxxxx=
br</A></I></B>=20
escreveu:</DIV>
<BLOCKQUOTE class=3Dreplbq=20
style=3D"PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #1010ff 2px =
solid">=C3=AF=C2=BB=C2=BF=20
<META content=3D"MSHTML 6.00.2900.3314" name=3DGENERATOR>
<STYLE></STYLE>
<DIV><FONT face=3DArial size=3D2>vou tentar,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>2^n - x^2 =3D c tal qque 1< n =
< 2007, como=20
todo n=C3=83=C2=BAmero pode ser expresso como diferen=C3=83=C2=A7a =
de dois quadrados, s=C3=83=C2=B3=20
existem "c" tal que n possa ser um quadrado, de sorte que c seja =
expresso=20
como diferen=C3=83=C2=A7a de dois quadrados</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2></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> Saturday, May 17, =
2008 11:02=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] OBM =
TERCEIRA FASE=20
=C3=A2=E2=82=AC=E2=80=9C N=C3=83=C2=8DVEL 3 -- 2=C3=82=C2=AA =
quest=C3=83=C2=A3o</DIV>
<DIV><BR></DIV><B><FONT face=3DArialNarrow-Bold>
<DIV align=3Dleft>XXIX OLIMP=C3=83=C2=8DADA BRASILEIRA DE =
MATEM=C3=83=C2=81TICA</DIV>
<DIV align=3Dleft>TERCEIRA FASE =C3=A2=E2=82=AC=E2=80=9C =
N=C3=83=C2=8DVEL 3 (Ensino M=C3=83=C2=A9dio)</DIV>
<DIV>PRIMEIRO DIA</DIV>
<DIV> </DIV><B><FONT face=3DArialNarrow-Bold size=3D2>
<DIV align=3Dleft>PROBLEMA 2</DIV></B></FONT><FONT =
face=3DTimesNewRoman=20
size=3D3>
<DIV>Para quantos n=C3=BAmeros inteiros c, - =
2007 <=3D c <=3D=20
2007 , existe um inteiro x tal que x^2 + c =C3=A9 m=C3=BAltiplo de =
2^2007?=20
</FONT></DIV>
<DIV><FONT face=3DTimesNewRoman size=3D3></FONT> </DIV>
<DIV><FONT face=3DTimesNewRoman size=3D3>algu=C3=A9m se =
habilita?</FONT></DIV>
<DIV><FONT face=3DTimesNewRoman size=3D3></FONT> </DIV>
<DIV><FONT face=3DTimesNewRoman size=3D3>grato, </FONT></DIV>
<DIV><FONT face=3DTimesNewRoman=20
=
size=3D3> &nbs=
p; =20
Douglas</DIV></FONT></B></FONT>
<DIV>
<HR SIZE=3D1>
Abra sua conta no <A=20
=
href=3D"http://br.rd.yahoo.com/mail/taglines/mail/*http://br.mail.yahoo.c=
om/">Yahoo!=20
Mail</A>, o =C3=83=C2=BAnico sem limite de espa=C3=83=C2=A7o para =
armazenamento!=20
</DIV></BLOCKQUOTE></BLOCKQUOTE><BR>
<P>
<HR SIZE=3D1>
Abra sua conta no <A=20
=
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om/">Yahoo!=20
Mail</A>, o =C3=BAnico sem limite de espa=C3=A7o para armazenamento!=20
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