New fractions:
$\dfrac{2+n}{3}; \dfrac{m+n}{4}; \dfrac{n+n}{6}$
So we have:
$\dfrac{2+n}{3}+\dfrac{m+n}{4}+\dfrac{2n}{6}=6$
$\Leftrightarrow \dfrac{n+2}{3}+\dfrac{m+n}{4}+\dfrac{n}{3}=6$
$\Leftrightarrow \dfrac{2n+2}{3}+\dfrac{m+n}{4}=6$
$\Leftrightarrow 4(2n+2)+3(m+n)=6.12$
$\Leftrightarrow 3m+11n=64$
$\Leftrightarrow m=\dfrac{64-11n}{3}$
$\forall n\in\mathbb{R}$, we can find out one value of $m$.
If $n, m\in\mathbb{N}$:
$64-11n>0\Leftrightarrow n<5,81$
$n=0\Rightarrow m=\dfrac{64}{3}$ (incorrect)
$n=1\Rightarrow m=\dfrac{53}{3}$ (incorrect)
$n=2\Rightarrow m=14$ (correct)
$n=3\Rightarrow m=\dfrac{31}{3}$ (incorrect)
$n=4\Rightarrow m=\dfrac{20}{3}$ (incorrect)
$n=5\Rightarrow m=3$ (correct)
$\Rightarrow (m;n)=(14; 2), (3;5)$
In conclusion: $m\times n=28$ or $15$.
