How Many Divisors?

Let us write $n = 2^{a} \times 3^{b} \times k$ where $k \in \mathbb{N}$ and the number of factors of $k$ is $c$ . From the statements given in the question we may obtain 2 equations: $56=(a+1)(b+2)c$ $70=(a+2)(b+2)c$ $(a+1)(b+2)c \ne 0$ therefore we can divide the second equation by the first to obtain $\dfrac{70}{56} = \dfrac{a+2}{a+1}$ yielding $a = 3$ . So, plugging this into the first equation we get $(b+2)c = 14$ . $b \ge 0$ and therefore $b+2 \ge 2$ . Since $(b+2)$ must be a factor of 14 and is at least 2, we get that $(b+2) \in [2,7,14]$ , each of which yields a solution. Therefore there are $\fbox{3}$ solutions