Question

Aratati ce nr N= $7^{n}$ · $9^{n}$ + $21^{n+1}$ · $3^{n}$ - 9 · $63^{n}$ este divizibil cu 13 pentru orice numar natural n

Answer 1

$N=7^n \cdot 9^n+21^{n+1} \cdot 3^n- 9 \cdot 63^n= \\ ~~~~=63^n+21 \cdot 21^n \cdot 3^n-9 \cdot 63^n= \\ ~~~~=63^n+21 \cdot 63^n-9 \cdot 63^n= \\ ~~~~=63^n(1+21-9)= \\ ~~~~=63^n \cdot13 \Rightarrow N ~ \vdots ~ 13.$