Question

Suma a patru nr naturale consecutive este cu 45 mai mare decat cel mai mare dintre nr .Aflati nr !

Answer 1

a+a+1+a+2+a+3=a+3+45
3a+3=45
3a=42
a=14
a+1=15
a+2=16
a+3=17

Answer 2

$x+y+z+t=t+45 \\ y=x+1 \\ z=y+1=x+2 \\ t=z+1=y+2=z+3 \\ x+x+1+x+2+x+3=x+3+45 \\ 4x+6=x+48 \\ 4x-x=48-6 \\ 3x=42 \\ x=42:3 \\ \boxed{x=14} \\ y=14+1 \\ \boxed{y=15} \\ z=15+1 \\ \boxed{z=16} \\ t=16+1 \\ \boxed{t=17}.$