20. Let $a, b$ and $c$ be positive integers such that $a^{2}=2 b^{3}=3 c^{5}$. What is the minimum possible number of factors of $a b c$ (including 1 and $a b c$ )?
20. 077 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 $a b c$, 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 $a^{2}=2 b^{3}=3 c^{5}$ we have $2^{2 p} 3^{2 q}=2^{3 r+1} 3^{3 s}=2^{5 v} 3^{5 w+1}$.
Considering indices of 2 we have $2 p=3 r+1=5 v$.
The smallest values of $(p, r, v)$ which satisfy this equation are $(5,3,2)$.
Considering indices of 3 we have $2 q=3 s=5 w+1$.
The smallest values of $(q, s, w)$ which satisfy this equation are $(3,2,1)$.
Using these values, $a b c=\left(2^{5} 3^{3}\right) \times\left(2^{3} 3^{2}\right) \times\left(2^{2} 3^{1}\right)=2^{10} 3^{6}$.
Any factors of $a b c$ will be of the form $2^{y} 3^{z}$ where $y \in\{0,1,2,3,4,5,6,7,8,9,10\}$ and $z \in\{0,1,2,3,4,5,6\}$.
There are 11 possibilities for $y$ and 7 possibilities for $z$. Hence $a b c$ has $11 \times 7=77$ factors.