Salutare!
[tex]\bf N=2^{n+3}\cdot3^{n} + 2^{n}\cdot3^{n+2}[/tex]
[tex]\bf N=2^{n}\cdot3^{n}( 2^{n+3-n} \cdot 3^{n-n}+ 2^{n-n}\cdot3^{n+2-n})[/tex]
[tex]\bf N=2^{n}\cdot3^{n}( 2^{3} \cdot 3^{0}+ 2^{0}\cdot3^{2})[/tex]
[tex]\bf N=2^{n}\cdot3^{n}\cdot(8+ 9)[/tex]
[tex]\bf N=2^{n}\cdot3^{n}\cdot 17[/tex] este divizibil cu 17