Notam n , n+1 , n+2 - aceste 3 numere
Din problema avem : [tex] \frac{n+n+1+n+2}{3} [/tex] +9 =720 ⇔[tex] \frac{n+n+1+n+2}{3} [/tex] = 720-9 ⇔ [tex] \frac{n+n+1+n+2}{3} [/tex] =711 ⇔ n + n+1 +n+2 =711*3 ⇔ 3n +3 =2133 ⇔ 3n =2130 ⇔n = 710
Acolo se putea da si factor comun : avand (3n+3)/3=711 si asta insemna ca 3(n+1)/3⇔ n+1=711 ⇔ n=710
