Din [tex] \frac{3}{4} < \frac{n}{6} =>n> \frac{3*6}{4} = \frac{18}{4} = \frac{9}{2} [/tex]
Din [tex] \frac{n}{6} < \frac{8}{9} => n< \frac{6*8}{9} = \frac{48}{9} = \frac{16}{3} [/tex]
Avem: [tex] \frac{9}{2} [/tex]<n< [tex] \frac{16}{3} [/tex] . n∈Z => n=5.