abc3 cu bara deasupra... si 3abc cu bara
abc nr in baza 10 rezulta a, b, c cifre a nenul
abc3-3abc=1000a+100b+10c+3-(3000+100a+10b+c)=1000(a-3)+100(b-a)+10(c-b)+(3-c)=
= 2000+10+6
= 1000(a-3)+100(b-a)+10(c-b-1)+(13-c)
13-c=6, c=7
c-b-1=1, b=5
b-a=0, a=5
5573-3557=2016
deci numarul abc=557
