[tex]\overline{a,bc}+\overline{c,ba}=8,28|\cdot 100\\
\overline{abc}+\overline{cba}=828\\
100a+10b+c+100c+10b+a=828\\
101a+20b+101c=828\\
101(a+c)=828-20b\\
Deorece\ membrul\ stang\ este\ multiplu\ de\ 101,atunci\ si\ membrul\\
drept\ trebuie\ sa\ fie\ multiplu\ de\ 101.Singura\ solutie\ convenabila\\
este\ b=1\\
101(a+c)=808\\
a+c=8;\ a,c\neq 0\\
S:\overline{abc}\in \{117,216,315,414,513,612,711\}[/tex]
