Three different natural numbers $x$, $y$ and $z$ each have exactly three natural-number factors. How many factors does $x^{2}y^{3}z^{4}$ have?
I notice that the problem is about the number of factors of natural numbers, which is related to their prime factorization. For example, $4$ has three factors: $1, 2, 4$, but $6$ has four factors: $1, 2, 3, 6$. So I can assume that $x = p^{2}$, $y = q^{2}$ and $z = r^{2}$, where $p, q, r$ are distinct prime numbers. Then, $x^{2}y^{3}z^{4} = p^{4}q^{6}r^{8}$. That is, if $n = p_{1}^{a_{1}}p_{2}^{a_{2}}\cdots p_{k}^{a_{k}}$, where $p_{1}, p_{2}, \ldots, p_{k}$ are distinct prime numbers and $a_{1}, a_{2}, \ldots, a_{k}$ are positive integers, then the number of factors of $n$ is $(a_{1} + 1)(a_{2} + 1)\cdots (a_{k} + 1)$. So, in my case, the number of factors of $x^{2}y^{3}z^{4}$ is $(4 + 1)(6 + 1)(8 + 1) = 5 \cdot 7 \cdot 9 = 315$.

