Question

Aflat 4 numere consecutive care are suma de 12 518

Answer 1

a-primul numar
a+1 -al doilea numar
a+2- al treilea numar
a+3-al patrulea numar

a+a+1+a+2+a+3=12518
4a+6=12518
4a=12518-6
4a=12512
a=12512/4
a=3128-primul nr.

a+1=3128+1=3129-al doilea nr.

a+2=3128+2=3130-al treilea nr.

a+3=3128+3=3131-al patrulea nr.

Proba:
3128+3129+3130+3131=12518

Answer 2

X=nr. necunoscut I nr.=x ll nr.=x+1 lll nr.=x+1+1 lVnr.=x+1+1+1 X+x+1+x+1+1+x+1+1+1=12518 4x+6=12518 4x=12518-6 4x=12512 x=12512:4 x=3128 l nr.=3128 ll nr.=3129 lll nr.=3130 lV nr.=3131