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suma a trei nr consecutive este219 aflati cele trei nr

Răspuns :

a+a+1+a+2=219
3a+3=219
3a=219-3
3a=216
a=216:3
a=72
Nr.sunt:72;73;74
Fie numerele:
1 numar- z
2 numar- z+1
3 numar- z+2 ⇒

z+z+1+z+2= 219
3z+3=219
3z=219-3
3z=216
z=216:3
z=72
II nr= z+1⇒ 72+1= 73
III nr= z+2⇒ 73+2=74