Question

1. Arata ca numarul N=5^2n x47+(25^2n+1)+90x6^n+6^n+1 este divizibil cu 24, pentru orice numar natural n.

Answer 1

[tex]N=5^{2n} *47+25^{n+1}+90*6^n+6^{n+1}\\ N= 5^{2n} *47+(5^2)^{n+1}+6^n(90+6)\\ N=5^{2n}(47+25)+6^n*96\\ N=72*5^{2n}+96*6^n\\ N=24(3*5^{2n}+4*6^n)\\ N\vdots24[/tex]