$a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have  following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value  of product $abc$.

Since $6\mid c$ we may write $c=2^r3^s$ with $r,s\ge 1$, so that $a^2=3c^5=2^{5r}3^{5s+1}$. Since $a^2$ is a square, $5r$ and $5s+1$ must be even. The minimal $r,s$ are then $r=2$ and $s=1$, so that $c=12$ and $a=864$. It follows that $b=72$, and
$$
abc=a^2=3b^3=2c^5.
$$
This value is minimal, because $c$ is minimal (see the comments above).
