Răspuns :
Fie n, n+1 si n+2-cele 3 nr. consecutive.
Ma=n+n+1+n+2/3=49
Ma=3n+3/3=49=>3n+3=49*3=147
3n+3=147
3n=147-3
3n=144
n=144:3
n=48
Numerele sunt:48, 49 si 50.
Ma=n+n+1+n+2/3=49
Ma=3n+3/3=49=>3n+3=49*3=147
3n+3=147
3n=147-3
3n=144
n=144:3
n=48
Numerele sunt:48, 49 si 50.