[tex]\displaystyle\\Din~\overline{abc}-\overline{cba}=\overline{xyz}~se~poate~deduce~ca~99(a-c)=\overline{xyz}~ceea~ce~implica~ca\\11|\overline{xyz}~si~9|\overline{xyz}~de~unde~11|(x+z-y)~si~9|(x+y+z)~ceea~ce~implica~ca\\x+y+z=9k,~k\in\{1,2,3\}~si~x+z-y~este~fie~11~fie~0.\\Daca~x+z-y~este~11~atunci~(x+z-y)+(x+y+z)=2(x+z)=11+9k~\\care~este~un~numar~par,~deci~k~trebuie~sa~fie~impar,~deci~fie~1~fie~3,~dar\\prin~verificare~directa~putem~deduce~ca~nu~avem~solutii~in~niciunul~din\\aceste~cazuri.\\\\[/tex]
[tex]Daca~x+z-y~este~0~atunci~(x+z-y)+(x+y+z)=2(x+z)=\\2y=9k~de~unde~k~trebuie~sa~fie~par,~adica~k=2.\\Deci~\boxed{a+b+c=x+y+z=18},deoarece~din~ipoteza~avem~ca:\\\{a,b,c\}=\{x,y,z\}.[/tex]
