From that statement we can create the following equation,
[tex]n\cdot \left(n+2\right)=48[/tex]
solving for n,
[tex]\begin{gathered} n^2+2n=48 \\ n^2+2n-48=0 \\ n_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\cdot \:1\cdot \left(-48\right)}}{2\cdot \:1} \\ n_{1,\:2}=\frac{-2\pm \:14}{2\cdot \:1} \\ n_1=\frac{-2+14}{2\cdot \:1},\:n_2=\frac{-2-14}{2\cdot \:1} \\ n=6,\:n=-8 \end{gathered}[/tex]
We can only use the positive number for this problem, therefore n = 6
From the above, the set of numbers is 6 and 6+2=8, since 6*8=48.
Answer: the greatest integer is 8
