Question

Suma a sase nr.este 399.Primele trei sunt consecutive pare.Urmatoarele trei sunt consecutive impare cu suma 303.Afla numerele.

Answer 1

$x+y+z+t+s+u=399 \\ y=x+2 \\ z=y+2=x+4 \\ s=t+2 \\ u=s+2=t+4 \\ t+s+u=303 \\ x+y+z=399-303 \\ x+y+z=96 \\ x+x+2+x+4=96 \\ 3x+6=96 \\ 3x=96-6=90 \\ x=\frac{90}{3} \\ \boxed{x=30} \\ y=x+2=30+2 \\ \boxed{y=32} \\ z=y+2=32+2 \\ \boxed{z=34} \\ t+s+u=303 \\ t+t+2+t+4=303 \\ 3t+6=303 \\ 3t=303-6 \\ 3t=297 \\ t=\frac{297}{3} \\ \boxed{t=99} \\ s=t+2=99+2 \\ \boxed{s=101} \\ u=s+2=101+2 \\ \boxed{s=103}$