xyzt=1000x+100y+10z+t=100(10x+y)+(10z+t)=100*xynumar+ztnumar
fie xy numar=a si ztnumar=b
atunci fractia devine
(100a+b)/(a+b) sa fie minim
(a+b+99a)/(a+b)=1+99a/(a+b)
1 constant, 99 constant⇒a cminim . a+b maxim⇒b maxim
a=10 c.m.mic nrde 2 cifre b=99, c.m.mare nr.de 2 cifre
abnumar=xyztnumar=1099
x=1;y=0; z=9 t=9
valoarea minima 1099/(10+99)=1099/109≈10,082...∈R
