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[SPAM] Re: [obm-l] Re:
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Desculpe Rafael, n=E3o estou conseguindo entender legal esse come=E7o:
escrever 99999899999 como 1e11 - 1e5 - 1, ao quadrado isso vale:
1e22 - 2e16 - 2e11 + 1e10 + 2e5 + 1. e da=ED para frente eu nem =
continuei ...tens como me ajudar ?
ge -----=20
From: Rafael Ando=20
To: obm-l@xxxxxxxxxxxxxx=20
Sent: Thursday, June 12, 2008 6:44 AM
Subject: [obm-l] Re:
1) Escreva 99999899999 como 1e11 - 1e5 - 1, ao quadrado isso vale:
1e22 - 2e16 - 2e11 + 1e10 + 2e5 + 1.
Vamos tentar entender essa expressao, fazendo operacao por operacao =
(desculpem pela falta de acentuacao, estou em um teclado que nao tem =
acentos e cedilha...): 1e22 - 2e16 sao 22-16-1=3D5 noves, seguido de um =
8, e 16 zeros (9999980000000000000000).
A proxima operacao eh fazer esse numero -2e11. O numero 8 =
transforma-se em 7, o zero na posicao 11 vira 8 e os 16-11-1 =3D 4 =
anteriores viram 9, os demais continuam 0. Entao temos o numero =
9999979999800000000000. As demais operacoes sao somas nos 0s, nao vai =
aparecer nenhum 9 e nem destruir os que ja temos.... entao a resposta eh =
9.
Letra B.
2) Podemos ver que 15x^2 eh congruente a (ie, termina com) 0 ou 5 =
mod10. y^2, com y inteiro, eh congruente a 0,1,4,5,6 ou 9, entao 7y^2 eh =
congruente a 0,7,8,5,2 ou 3. Pra resposta ser 9, teriamos que ter 7y^2 =
congruente a 1 (para o caso 15x^2 conguente a 0) ou 6 (para o congruente =
a 5), nenhum deles eh possivel. Logo, essa equacao nao tem solucao =
(diofantina).
Letra A.
On Mon, Apr 7, 2008 at 5:47 AM, fgb1 <fgb1@xxxxxxxxxxxx> wrote:
Amigos... T=F5 precisando de ajuda em 2 problemas:
1) Quantos "noves" existem na representa=E7=E3o decimal de =
99999899999^2?
a) 7
b) 9
c) 11
d) 13
e) 15
2) Quantos inteiros x e y existem tais que 15x^2 - 7y^2 =3D 9/
a) 0
b) 1
c) 2
d) 3
e) 4
Desde j=E1 agrade=E7o
=
=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=
=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=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Instru=E7=F5es para entrar na lista, sair da lista e usar a lista em
http://www.mat.puc-rio.br/~obmlistas/obm-l.html
=
=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=
=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=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
--=20
Rafael
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<META http-equiv=3DContent-Type content=3D"text/html; =
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<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Desculpe Rafael, n=E3o estou =
conseguindo entender=20
legal esse come=E7o:</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"><STRONG><FONT color=3D#ff0000>escrever =
99999899999=20
como 1e11 - 1e5 - 1, ao quadrado isso vale:<BR><BR>1e22 - 2e16 - 2e11 =
+ 1e10 +=20
2e5 + 1.</FONT> e da=ED para frente eu nem continuei =
...tens como me=20
ajudar ?<BR></STRONG><BR>ge ----- </DIV>
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Drafael.ando@xxxxxxxxx =
href=3D"mailto:rafael.ando@xxxxxxxxx";>Rafael=20
Ando</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, June 12, 2008 =
6:44=20
AM</DIV>
<DIV style=3D"FONT: 10pt arial"><B>Subject:</B> [obm-l] Re:</DIV>
<DIV><BR></DIV>1) Escreva 99999899999 como 1e11 - 1e5 - 1, ao quadrado =
isso=20
vale:<BR><BR>1e22 - 2e16 - 2e11 + 1e10 + 2e5 + 1.<BR><BR>Vamos tentar =
entender=20
essa expressao, fazendo operacao por operacao (desculpem pela falta de =
acentuacao, estou em um teclado que nao tem acentos e cedilha...): =
1e22 - 2e16=20
sao 22-16-1=3D5 noves, seguido de um 8, e 16 zeros=20
(9999980000000000000000).<BR><BR>A proxima operacao eh fazer esse =
numero=20
-2e11. O numero 8 transforma-se em 7, o zero na posicao 11 vira 8 e os =
16-11-1=20
=3D 4 anteriores viram 9, os demais continuam 0. Entao temos o numero=20
9999979999800000000000. As demais operacoes sao somas nos 0s, nao vai =
aparecer=20
nenhum 9 e nem destruir os que ja temos.... entao a resposta eh=20
9.<BR><BR>Letra B.<BR><BR>2) Podemos ver que 15x^2 eh congruente a =
(ie,=20
termina com) 0 ou 5 mod10. y^2, com y inteiro, eh congruente a =
0,1,4,5,6 ou 9,=20
entao 7y^2 eh congruente a 0,7,8,5,2 ou 3. Pra resposta ser 9, =
teriamos que=20
ter 7y^2 congruente a 1 (para o caso 15x^2 conguente a 0) ou 6 (para o =
congruente a 5), nenhum deles eh possivel. Logo, essa equacao nao tem =
solucao=20
(diofantina).<BR><BR>Letra A.<BR><BR>
<DIV class=3Dgmail_quote>On Mon, Apr 7, 2008 at 5:47 AM, fgb1 <<A=20
href=3D"mailto:fgb1@xxxxxxxxxxxx";>fgb1@xxxxxxxxxxxx</A>> wrote:<BR>
<BLOCKQUOTE class=3Dgmail_quote=20
style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: =
rgb(204,204,204) 1px solid">Amigos...=20
T=F5 precisando de ajuda em 2 problemas:<BR><BR>1) Quantos "noves" =
existem na=20
representa=E7=E3o decimal de 99999899999^2?<BR><BR>a) 7<BR>b) =
9<BR>c) 11<BR>d)=20
13<BR>e) 15<BR><BR>2) Quantos inteiros x e y existem tais que 15x^2 =
- 7y^2 =3D=20
9/<BR><BR>a) 0<BR>b) 1<BR>c) 2<BR>d) 3<BR>e) 4<BR><BR>Desde j=E1=20
=
agrade=E7o<BR><BR><BR>=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=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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D<BR>Instru=E7=F5es=20
para entrar na lista, sair da lista e usar a lista em<BR><A=20
href=3D"http://www.mat.puc-rio.br/%7Eobmlistas/obm-l.html"=20
=
target=3D_blank>http://www.mat.puc-rio.br/~obmlistas/obm-l.html</A><BR>=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=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=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D<BR></B=
LOCKQUOTE></DIV><BR><BR=20
clear=3Dall><BR>-- <BR>Rafael </BLOCKQUOTE></BODY></HTML>
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