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Concursul "Congruente", problema 3
Posted: Sun Nov 09, 2008 11:06 pm
by Claudiu Mindrila
Aratati ca nu exista numere de forma \( \overline{abc} \) care verifica relatia \( \overline{ab}^2-\overline{bc}^2=\overline{abc}. \)
Nicolae Stanica, Braila
Posted: Mon May 04, 2009 6:25 pm
by salazar
\( (10a+b)^2-(10b+c)^2=100a+10b+c \)
\( 100a^2+20ab+b^2-b^2-20bc-c^2=100a+10b+c \)
\( 100a^2-100a+20ab-10b=20bc+c^2+c \)
\( 100a(a-1)+10b(2a-1)=20bc+c(c+1) \)
-din cele spuse mai sus\( \Longrightarrow c(c+1)\vdots 10\Longrightarrow c\in\lbrace0,4,5,9\rbrace \)
-c=0\( \overline{ab}^2-\overline{b0}^2=\overline{ab0}\Longrightarrow U(b)=0\Longrightarrow b=0 \)FALS
-c=4 \( U(b)=0\Longrightarrow b=0 \)FALS
-c=5 \( U(b)=0\Longrightarrow b=0 \)FALS
-c=9 \( U(b)=0\Longrightarrow b=0 \)FALS