[tex]\displaystyle\\
8)\\
5^2 = 25\\
5^3 = 125\\
\text{Numerele pare cuprinse intre 25 si 125 sunt:}\\
26;~28;~30;~32;~\hdots ~;~124\\
\text{Numarul de numere pare cuprinse intre 25 si 125 este:}\\\\
n = \frac{124-26}{2}+1 = \frac{98}{2}+1 = 49+1=50~\text{de numere pare}\\\\
\text{Suma numerelor pare cuprinse intre 25 si 125 este:}\\\\
26+28+30+32+~\cdots~+124 = \frac{50(124+26)}{2}=25\times 150=\boxed{\bf 3750}[/tex]
[tex]\displaystyle\\9)\\
14\cdot10^n\in\{14;140;1400;14000;~...~;14000...000;...\}\\
\text{deoarece }~n\in\{0;1;2;3;...\}\\\\
\text{Suma cifrelor lui }~14\cdot10^n\text{ este: }~S=1+4+n\cdot0= \boxed{5}\\
\text{D\! \!efinitie: }\\
_\texttt{\bf Suma a 2 numere este divizibila cu 9 daca suma tuturor cifrelor }\\
_\texttt{\bf de la ambele numere este divizibila cu 9}\\
S(14\cdot 10^n +337 ) = (1 + 4) + (3+3+7) = 18~\vdots~9\\
\Longrightarrow~\boxed{x=(14\cdot 10^n +337 )~\vdots~9}[/tex]
[tex]\displaystyle\bf\\10)\\
\text{Numerele sunt:}\\
\overline{abc};~\overline{acb};~\overline{bac};~\overline{bca};~\overline{cab};~\overline{cba}\\
S=\overline{abc}+\overline{acb}+\overline{bac}+\overline{bca}+ \overline{cab}+\overline{cba}=\\
=100a+10b+c+100a+10c+b +100b+10a+c+\\
+100b+10c+a+100c+10a+b+100c+10b+a =\\
=100(a+a+b+b+c+c)+10(a+a+b+b+c+c)+\\
+(a+a+b+b+c+c)=\\
=100(2a+2b+2c)+10(2a+2b+2c)+(2a+2b+2c)=\\
200(a+b+c)+20(a+b+c)+2(a+b+c)=\\
(200+20+2)(a+b+c)=\boxed{\bf222(a+b+c)~\vdots~(a+b+c)}[/tex]
