Question

impartind suma a trei numere consecutive la 8, obtin catul 108 si restul 3. care sunt numerele

Answer 1

(a+a+1+a+2):8=108 rest 3
(3a+3):8=108 rest 3
3a+3=108×8+3
3a+3=864+3
3a+3=867
3a=867-3
3a=864
a=864:3
a=288
scuze am scris18 in loc de 108!Am corectat!
Numerele sunt:288,289,290.

Answer 2

(a+b+c):8=108 rest 3
a+b+c=108x8+3
a+b+c=867
a=a
b=a+1
c=a+2
a+a+1+a+2=867
ax3+3=867
ax3=867-3
ax3=864
a=864:3
a=288
b=289
c=290
V:288+289+290=867
(288+289+290):8=108 rest 3
867:8=108 rest 3
108 rest 3=108 rest 3
Sper ca te-am ajutat!!!