Deoarece profesorul imparte problemele in mod egal elevilor deducem ca
[tex] \frac{3n+9}{2n+2} \in N=> \frac{2(3n+9)}{2n+2} \in N=> \frac{6n+18}{2n+2} \in N=>\\
\frac{6(n+1)+12}{2(n+1)} \in N=> \frac{6(n+1)}{2(n+1)}+\frac{12}{2(n+1)} \in N=>\\
3+\frac{12}{2(n+1)} \in N=>\frac{6}{n+1} \in N=>n+1\in D_6=\{1,2,3,6\}=>\\
n\in\{0,1,2,5\}[/tex]
Pentru n=5=>Sunt 2·5+2=12 elevi si 3·5+9=24 probleme.
