$\prod\limits_{i=1}^n \frac{3a_i+2}{2a_i+1}, a_i\geq 1$ $$\prod\limits_{i=1}^n \frac{3a_i+2}{2a_i+1},
a_i\geq 1$$
Claim: This product is never an integer ($a_i$ integer).
 A: It's false:
$$
\frac{29}{19}\frac{38}{25}\frac{44}{29}\frac{83}{55}\frac{125}{83}=8
$$
A: Some more counterexamples: 
$$ \frac{20}{13} \frac{ 38}{25} \frac{ 83}{55} \frac{ 125}{83} \frac{143}{95} = 8$$
$$\frac{8}{5}{\frac {65}{43}}{\frac {74}{49}}{\frac {83}{55}}{\frac {98}{65
}}{\frac {125}{83}}{\frac {143}{95}}{\frac {209}{139}}{\frac {215}
{143}}{\frac {278}{185}}=64$$
$${\frac {11}{7}}{\frac {38}{25}}{\frac {65}{43}}{\frac {83}{55}}{
\frac {98}{65}}{\frac {125}{83}}{\frac {143}{95}}{\frac {215}{143}}
=28$$
$${\frac {17}{11}}{\frac {20}{13}}{\frac {26}{17}}{\frac {74}{49}}{
\frac {83}{55}}{\frac {98}{65}}{\frac {125}{83}}{\frac {143}{95}}{
\frac {209}{139}}{\frac {278}{185}}=64$$
EDIT:
In general, given a finite set of fractions $f_j$, consider the set $P$ of primes
that divide their denominators.  If the $p$-adic order of $f_j$ is $\nu_p(f_j)$,
the condition for $\prod_j f_j$ to be an integer is $\sum_j \nu_p(f_j) \ge 0$
for all $p \in P$.  Let $x_j$ be binary variables (i.e. possible values $0$ and $1$).  Then $\prod_j f_j^{x_j}$ is an integer iff $\sum_j \nu_p(f_j)\; x_j \ge 0$ for all $p \in P$.  So what we want to do is find  solutions to the
system of linear inequalities
$$ \sum_j \nu_p(f_j)\; x_j \ge 0, \ p \in P $$
$$ \sum_j x_j \ge 1 $$
$$ x_j \in \{0,1\}$$
Integer linear programming software can be used for this (I used the Optimization package in Maple).
