[tex]\it Not\breve{a}m \ \overline{abc} = 3n\ \ \ \ (*)
\\\;\\
\overline{abc00} = 100\cdot\overline{abc} \stackrel{(*)}{\Longrightarrow } \overline{abc00} = 100\cdot 3n= 300n[/tex]
Egalitatea din enunț devine:
[tex]\it 3+6+9+\ ...\ +3n =300n \Rightarrow 3(1+2+3+\ ...\ +n) =300n \Rightarrow
\\\;\\ \\\;\\
\Rightarrow 3\cdot\dfrac{n(n+1)}{2}=300n|_{:3n} \Rightarrow \dfrac{n+1} {2}=100 \Rightarrow n+1=200 \Rightarrow
\\\;\\ \\\;\\
\Rightarrow n=199 \stackrel{(*)}{\Longrightarrow} \overline{abc} =3\cdot199 =597 \Longrightarrow u(\overline{abc}) = 7[/tex]
