abc=100*a+10*b+c
bca=100*b+10*c+a
cab=100*c+a*10+b
inlocuiesti in ecuatie: 100a+10a+a+100c+10c+c+100b+10b+b=555
111a+111b+111c=555
Dai factor comun 111(a+b+c)=555=> a+b+c=5
dam valori pentru a, b ,c
a=1=>b+c=4=>b=1,c=3=> abc=113,bca=131,cab=311 si 113+131+311=555
(nu s-a specificat daca numerele sunt distincte deci putem avea a=b sau a=c sau c=b) b=2,c=2=> abc=122,bca=221,cab=212si 122+221+212=555
b=3.c=1=> abc=131,bca=311,cab=131 131+131+311=555
a=2=> b+c=3=>b=1,c=2 abc=212,bca=122,cab=221 221+122+212=555
b=2,c=1
a=3=>b+c=2=>b=1,c=1
Se observa ca volorile se repeta .
Sper ca ai inteles!
Bafta!
