[tex]n=3^{27}+3\cdot 2^{36}+15\cdot 3^{24}+2^{26}\cdot 3^{26}=\\
=3^{25}\cdot 3^{2}+5\cdot 3^{25}+3^{25}\cdot 3\cdot 2^{26}+3\cdot 2^{36}=\\
=3^{25}(3^2+5+3\cdot 2^{26})+3\cdot 2^{36}\\
3\cdot2^{26}<3^{25}<=>2^{26}<3^{24}<=>(2^{13})^2<(3^{12})^2<>8192^2<531441^2[/tex]
Aplicand teorema impartirii cu rest deducem ca restul impartirii lui n la [tex]3^{25}[/tex] este [tex]3\cdot 2^{36}.[/tex]
