$a^2 = 2b^3 = 3c^5$ 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 $abc$.
 A: 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).
A: Since $a^2=2b^3$, this means $a = \sqrt{2b} \cdot b$. Hence, $b=2l^2$. This gives us $a=4l^3$.
We also need $a^2 = 3c^5 \implies 16l^6 = 3c^5 \implies c = \left(\dfrac{2^4l}3\right)^{1/5} \cdot l$. This means $l = 3\cdot 2 \cdot m^5$.
Hence, we have $(a,b,c) = (864m^{15},72m^{10},12m^6)$. The minimum is obtained when $m=1$. Hence, $(a,b,c) = (864,72,12)$.
A: Introducing any prime factors apart from $2$ and $3$ makes things worse. Therefore we may assume $a=2^\alpha3^\beta$, which leads to
$$b^3=2^{2\alpha-1}3^{2\beta},\qquad c^5=2^{2\alpha}3^{2\beta-1}\ .$$
therefore we have to find the smallest $\alpha\geq0$, $\beta\geq0$ such that
$$2\alpha-1=0\quad(3),\qquad 2\alpha=0\quad(5)\ ,$$
and
$$2\beta=0\quad(3),\qquad2\beta-1=0\quad(5)\ .$$
The obvious solutions are $\alpha=5$, $\>\beta=3$, and lead to the triple $(a,b,c)=(864,72,12)$, so that the minimal product is $abc=746\,496$. 
