Balance the Factors

It is clear that each of $a$ , $b$ and $c$ must have prime factors including 2 and 3 and, since we are seeking a minimal number of factors of $abc$ , these must be the only prime factors.

Let $a = 2^p 3^q$ , $b = 2^r 3^s$ and $c = 2^v 3^w$ where $p$ , $q$ , $r$ , $s$ , $v$ and $w$ are positive integers or zero. Since $a^2 = 2b^3 = 3c^5$ we have $2^{2p} 3^{2q} = 2^{3r+1} 3^{3s} = 2^{5v} 3^{5w+1}$ .

Considering indices of 2 we have $2p = 3r+1 = 5v$ . The smallest values of $(p, r, v)$ which satisfy this equation are $(5, 3, 2)$ .

Considering indices of 3 we have $2q = 3s = 5w+1$ . The smallest values of $(q, s, w)$ which satisfy this equation are $(3, 2, 1)$ .

Using these values, $abc = (2^5 3^3) \times (2^3 3^2) \times (2^2 3^1) = 2^{10} 3^6$ .

Any factors of $abc$ will be of the form $2^y 3^z$ where $y \in \{0, 1, \dots , 10\}$ and $z \in \{0, 1, \dots , 6\}$ . There are 11 possibilities for $y$ and 7 possibilities for $z$ . Hence $abc$ has $11 \times 7 = 77$ factors.