abcd+abc+ab+a=2222
a,b,c,d<10;
abcd=1000a+100b+10c+d
=> (1000a+100b+10c+d)+(100a+10b+c)+(10a+b)+a=2222
1111a + 111b + 11c +d =2222
a=2; 1111•2=2222 => b=c=d=0
=> abcd=2000
2000+200+20+2=2222
a=1 nu convine;
1111+111b+11c+d=2222, suma 111b+11c+d nu poate fi 1111, chiar daca b,c si d ar avea valori maxime (999+99+9=1007)
