[tex]1.~Cum~ \overline{ab}~si~ \overline{ba}~sunt~simultan~prime,~rezulta~ca~a,b \in \{1,3,7,9\}. \\ \\ Pentru~a=1~se~obtine~b \in \{3,7\}. \\ \\ Pentru~a=3~se~obtine~b \in \{1,7\}. \\ \\ Pentru~a=7~se~obtine~b \in \{1,3,9\}. \\ \\ Pentru~a=9~se~obtine~b=7. \\ \\ Observatie:~Pentru~rezolvarea~acestui~exercitiu,~trebuie~cunoscute \\ \\ numerele~prime~de~doua~cifre(11,~13,~17,~19,~23,~29,~31,~37,~41,~43, \\ 47,~53,~59,~61,~67,~71,~73,~79,~83,~89,~97).[/tex]
[tex]2.~Avem:~ \overline{ab}+ \overline{ba}=(10a+b)+(10b+a)=11a+11b=11(a+b). \\ \\ Deci~11(a+b)~este~patrat~perfect,~de~unde~rezulta~ca~a+b=11k^2 \\ \\ (k \in~N).~dar~2 \leq a+b \leq 18,~\Leftrightarrow 2 \leq 11k^2 \leq 18,~deci~k=1. \\ \\ Asadar~a+b=11. \\ \\ \underline{Solutie}: \overline{ab} \in \{29,38,47,56,65,74,83,92\}.[/tex]
[tex]3.~ \overline{abcabc}=1000 \cdot \overline{abc}+ \overline{abc}= \overline{abc} \cdot 1001=7 \cdot 11 \cdot 13 \cdot \overline{abc}. \\ \\ Descompunerea~acestui~numar~in~produs~de~factori~primi~este~deci \\ \\ 7 \cdot 11 \cdot 13 \cdot \overline{abc}.~Fiecare~numar~prim~este~la~puterea~1. \\ \\ Deci~numarul~are~(1+1)(1+1)(1+1)(1+1)=16~divizori.[/tex]
