Question

Aratati ca numerele de forma 72·$12^{n}$+$3^{n+3}$·$4^{n+2}$ sunt divizibile cu 63, unde n∈N*$4^{n+2}$

Answer 1

   
[tex]72\cdot 12^n +3^{n+3} \cdot 4^{n+2} = \\ = 72\cdot 12^n +3^n\cdot3^3 \cdot 4^n\cdot4^2 = \\ = 72\cdot 12^n +3^n\cdot 4^n\cdot3^3 \cdot4^2 = \\ = 72\cdot 12^n +(3\cdot 4)^n\cdot3^3 \cdot4^2 = \\ = 72\cdot 12^n +12^n\cdot3^3 \cdot4^2 = \\ = 72\cdot 12^n +12^n\cdot3^3 \cdot4^2 = \\ = 12^n(72+3^3 \cdot4^2)= \\ = 12^n(8\cdot 9+4^2\cdot3^3)= \\ = 12^n(2^3\cdot 3^2+(2^2)^2\cdot3^3)= \\ = 12^n(2^3\cdot 3^2+2^4 \cdot 3^3)= \\ = 12^n \cdot (2^3\cdot 3^2)(1+2^1 \cdot 3^1)= [/tex]

[tex]= 12^n \cdot (2^3\cdot 3^2)(1+2 \cdot 3)= \\ = 12^n \cdot (2^3\cdot 3^2)(1+6)= \\ = 12^n \cdot 2^3\cdot 3^2 \cdot 7= \\ = 12^n \cdot 2^3\cdot (9 \cdot 7)= \\ = \boxed{12^n \cdot 2^3\cdot 63 ~\vdots~63} [/tex]