[tex]1) \\ a=1+2+3+.....+100+51\cdot 101= \\ = \frac{100(100+1)}{2} +51\cdot 101= \\ = 50\cdot 101+51\cdot 101=101(50+51)=\boxed{101\cdot101~~~patrat ~perfect} \\ \\ 2) (Compunere ~problema) \\ \text{Sa se demonstreze ca din n numere consecutive, } \\ \text{exista un numar divizibil cu n, ori care ar fi n si } \\ \text{oricare ar fi numerele consecutive.}[/tex]
[tex]3) \\ \frac{7+28+37}{3} =\frac{72}{3} =\boxed{24} \\ \\ 4) \\ 15\cdot 12\cdot 3+12\cdot 7\cdot 3\cdot 13 = 12\cdot 3(15 + 7 \cdot 13) \\ =\ \textgreater \ \;\; Factorul ~comun~ este \boxed{12\cdot 3}=\boxed{36} \\ \\ 5) \\ n~\vdots ~5 daca~ultima~cifra~a~numarului~este:~ \boxed{0~sau~5} \\ \\ 6) \\ D_{36} = \boxed{\{ 1;~2;~3;~4;~6;~9;~12;~18;~36 \}}[/tex]
