Enunțul nu e foarte clar. Vom reformula:
[tex]\bf Se\ consider\breve a\ num\breve arul\ \overline{ab},\ cu\ a\ne b\ . \ \ Media\ aritmetic\breve a\ ponderat\breve a\\ \\ a\ cifrelor\ sale,\ cu\ ponderile\ 1\ \d si\ 6,\ este\ egal\breve a\ cu\ 2\ .\\ \\ Ar\breve ata\c{\bf t}i\ c\breve a\ \sqrt{\overline{ab}}\ este\ p\breve atratul\ unui\ num\breve ar\ natural\ .[/tex]
Rezolvare:
[tex]\it m_{a.p.}=\dfrac{a+6b}{7}=2 \Rightarrow a+6b=14 \Rightarrow 6b=14-a \in M_6 \Rightarrow a=8,\ \ b=1\\ \\ \\ \sqrt{\overline{ab}}=\sqrt{81}=9=3^2[/tex]
