Question

Folosind proprietatile adunarii nr naturale calculati a) 2301 + 3698 + 689 + 302 b)[(324+76+100] + (109+16+1)

Answer 1

a) 2301+3968+689+302=
(2301+689)+(3698+302)=
2990+4000= 6990

b) (324+76+100)+(109+16+1)=
(400+100)+(109+1+16)=
500+(110+16)=
500+126=626

Answer 2

(2301+689)+(3698+302)=2990+4000=6990

(324+76)+(109+1)+(100+16)=400+110+116=626