Friend gave me this problem (although I don't think he knows the answer). The question is whether the product $$ \frac{3m_1+2}{2m_1+1}\cdot\frac{3m_2+2}{2m_2+1}\cdots \frac{3m_n+2}{2m_n+1} $$ can be an integer for some distinct $m_i \in \mathbb{N}$ (here $0 \not\in \mathbb{N}$, otherwise we would have simple solution $n=1, m_1=0$, giving product $2$). So basically this is about products of combinations of numbers $$ \frac{5}{3},\frac{8}{5},\frac{11}{7},\frac{14}{9}, \frac{17}{11}, \frac{20}{13}, \frac{23}{15}, \frac{26}{17}, \frac{29}{19}, \frac{32}{21}, \dots $$
My try so far:
We can rule out fractions which have denominator divisible by $3$, since no numerator can be divisible by $3$, so basically $m \not\equiv 1 \pmod 3$. I have then tried computer checking all products with $m_i \in \{2,3,5,6,8,9,\dots,39\}$, none of them yield an integer.
There are solutions for many values of $n$. You can find here solutions for $n=5,\;\;7,8,9,10,\ldots, 21$.
One of solutions for $n=5$:
$$ (m_1,m_2,m_3,m_4,m_5)=(9,12,14,27,41): $$
$$ \dfrac{3\cdot 9+2}{2\cdot 9 + 1} \times \dfrac{3\cdot 12+2}{2\cdot 12 + 1} \times \dfrac{3\cdot 14+2}{2\cdot 14 + 1} \times \dfrac{3\cdot 27+2}{2\cdot 27 + 1} \times \dfrac{3\cdot 41+2}{2\cdot 41 + 1} = \\ \dfrac{\color{red}{29}}{\color{violet}{19}} \times \dfrac{\color{violet}{38}}{\color{blue}{25}} \times \dfrac{44}{\color{red}{29}} \times \dfrac{\color{green}{83}}{55} \times \dfrac{\color{blue}{125}}{\color{green}{83}} = \\ \dfrac{503063000}{62882875} = 8.$$
Another examples of solutions for $n=5$:
$(9,12,14,21,71) \rightarrow 8$;
$(6,12,27,41,47) \rightarrow 8$;
$(6,12,21,47,71) \rightarrow 8$;
$(9, 11, 12, 30, 101) \rightarrow 8$;
$(8, 9, 14, 71, 107) \rightarrow 8$;
$(6, 8, 47, 71, 107) \rightarrow 8$;
$(5, 12, 41, 47, 110) \rightarrow 8$;
$(6, 8, 39, 87, 131) \rightarrow 8$;
$(6, 11, 17, 99, 132) \rightarrow 8$;
$(3, 99, 123, 126, 132) \rightarrow 8$;
$\ldots$
$(3,30,1943,5351,5652) \rightarrow 8$;
$(3,30,2081,4371,5912) \rightarrow 8$;
$(3,30,2133,4680,5103) \rightarrow 8$;
$\ldots$
Now focus on $n\ge 7$. For each $n\ge 7$ we search integer solution $q, (m_1,m_2,\ldots,m_n)$ for equation $$ \prod_{j=1}^n \dfrac{3m_j+2}{2m_j+1} = q \tag{*}. $$
For each $n\ge 7$ there are many solutions of eq. $(*)$. Few examples with rather small values $m_j$:
Note 1: if $n\ge 8$, then eq. $(*)$ has solutions for different integer $q$.
Note 2: this integer product is divisible by $2$, $4$, $8$ in many cases (depends on parity of numbers $m_j$); solutions with odd $q$ exist too, but are much more rare.