Răspuns:
Explicație pas cu pas:
[tex]\bf \overline{abc}~~\vdots~3[/tex]
(a + b + c) ⋮ 3
[tex]\bf \overline{abc}=100a +10b+c[/tex]
[tex]\bf a +99a +b +9b + c =[/tex]
[tex]\bf (a +b+c)+99a +9b =[/tex]
[tex]\purple{\boxed{\bf ~[(a +b+c)+9\cdot (11a +b)] ~~\vdots ~~3~}}[/tex]
↓ ↓
⋮ 3 ⋮ 3
Sau
(a + b + c) ⋮ 3
si
9·(11a + b) ⋮ 3 ⇒ [(a + b + c) + 9·(11a + b)] ⋮ 3
==pav38==
