[tex]\displaystyle\\A=10^n+38=\\\\=1\!\underbrace{000.....0000}_{\text{\bf n zerouri}}+38\\\\=1\!\underbrace{000.....00}_{\text{\bf n-2 zerouri}}\!\!38\\\\\text{Suma cifrelor acestui numar este:}\\\\S=1+0+0+0+....+0+0+0+3+8=1+3+8=4+8=12\\\\12~\vdots~3\\\\\implies~~\boxed{1\!\underbrace{000.....00}_{\text{\bf n-2 zerouri}}\!\!38~\vdots~3}\\\\\\\texttt{Cazuri particulare:}\\\\n=0~\implies~A=10^n+38=10^0+38=1+38=39~\vdots~3\\\\n=1~\implies~A=10^n+38=10^1+38=10+38=48~\vdots~3\\[/tex]
A=10ⁿ+38 ,unde n∈N;
Un numar natural este divizibil cu 3 daca si numai daca suma cifrelor acestuia este divizibila cu 3, deci in cazul nostru, obserbam ca pentru n >=1 => 10ⁿ=100...00 (de n ori) => S.c(10ⁿ+38)=1+38=39 => 3/39 (adevarat) => 3/A; iar pentru n=0 => 10ⁿ+38=1+38=39 => S.c(39)=3+9=12 => 3/12 (adevarat) => 3/A.
In concluzie, pentru orice n∈N, A=10ⁿ+38 este divizibil cu 3.
