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Concursul "Congruente", problema 3

 
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Claudiu Mindrila
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Joined: 01 Oct 2007
Posts: 557
Location: Targoviste

PostPosted: Sun Nov 09, 2008 11:06 pm    Post subject: Concursul "Congruente", problema 3 Reply with quote

Aratati ca nu exista numere de forma \overline{abc} care verifica relatia \overline{ab}^2-\overline{bc}^2=\overline{abc}.

Nicolae Stanica, Braila
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elev, clasa a XI-a, C. N. "C-tin Carabella", Targoviste
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salazar
Thales


Joined: 06 Apr 2009
Posts: 100
Location: Alba Iulia

PostPosted: Mon May 04, 2009 6:25 pm    Post subject: Reply with quote

(10a+b)^2-(10b+c)^2=100a+10b+c
100a^2+20ab+b^2-b^2-20bc-c^2=100a+10b+c
100a^2-100a+20ab-10b=20bc+c^2+c
100a(a-1)+10b(2a-1)=20bc+c(c+1)
-din cele spuse mai sus\Longrightarrow c(c+1)\vdots 10\Longrightarrow c\in\lbrace0,4,5,9\rbrace
-c=0\overline{ab}^2-\overline{b0}^2=\overline{ab0}\Longrightarrow U(b)=0\Longrightarrow b=0FALS
-c=4 U(b)=0\Longrightarrow b=0FALS
-c=5 U(b)=0\Longrightarrow b=0FALS
-c=9 U(b)=0\Longrightarrow b=0FALS
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