[tex]\overbar{abc}=\overbar{ab}+\overbar{bc}+\overbar{ca}\\
\\ 100a+10b+c=10a+b+10b+c+10c+a\\
\\ 100a+10b+c=11a+11b+11c~|-11a-10b-c\\
\\ 89a=b+10c\\
\\89a=\overbar{bc}\\
\\ 10 \leq \overbar{bc} \leq 99\\
\\ (1 \leq a \leq 9~|\cdot 89)\\
\\89 \leq 89a \leq 801\\
\\ In~acest~caz~89a=\{89,~178,~...,~801/}~Dar~89a~este~egal~cu~un~nr~de~doua~cifre~(bc)\\
\\ \Rightarrow 89a=\overbar{bc}~<=>\\
\\ <=>~a=1,~si~\overbar{bc}=89 \Rightarrow b=8,~c=9[/tex]
Raspuns: a+b+c=1+8+9=18
