Question

sa se afle trei numere naturale consecutive,stiind ca daca le adunam cu tripul sumei lor,obtimen 108

Answer 1

[tex]a+1+a+2+a+3(a+1+a+2+a)=108\\ a+a+1+a+2+3a+3+3a+6+3a=108\\ 12a+12=108\\ 12a=96\\ a=8.\\ a+1=9\\ a+2=10\\\\ Proba:8+9+10+3(8+9+10)=108\\ 27+81=108\\ 108=108[/tex]

Answer 2

X+3X=108
X=108:4=27
A+B+C=27
a+a+1+a+2=27
3a=27-3=28
a=24:3=8
b=8+1=9
c=8+2=10