Suppose that $x,y,z$ are positive integers satisfying $x \le y \le z$, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of $z$?
Since this is for only positive integers, and there are sums and products involved, I think that this can be approached using Simon's Favorite Factoring Trick. I am not sure how though. Help is greatly appreciated
For those who do not know what Simon's Favorite Factoring Trick is, it is a method of factoring by grouping. For example, say that we want to factor $xy+x+y+1$. We can factor this as so: $$xy+x+y+1$$ $$x(y+1)+y+1$$ $$x(y+1)+1(y+1)$$ $$(x+1)(y+1)$$
Suppose $x,y,z$ are positive integers, with $x \le y \le z$, such that $xyz=2(x+y+z)$. \begin{align*} \text{Then}\;\;&xyz=2(x+y+z)\\[4pt] \implies\;&xyz \le 2(3z)\\[4pt] \implies\;&xy \le 6\\[4pt] \implies\;&x^2 \le 6\\[4pt] \implies\;&x \le 2\\[4pt] \end{align*} Consider two cases . . .
Case $(1)$:$\;x=1$. \begin{align*} \text{Then}\;\;&xyz=2(x+y+z)\\[4pt] \iff\;&yz =2(1+y+z)\\[4pt] \iff\;&yz - 2y -2z - 2 = 0\\[4pt] \iff\;&(y-2)(z-2)=6 \end{align*} which leads to a small number of possibilities for $y,z$, left for you to complete.
Case $(2)$:$\;x=2$.
Since $xy\le 6$, and $x \le y$, we get $2\le y\le 3$.
Using $x=2$, for each of $y=2,y=3$, solve the equation $xyz=2(x+y+z)$ for $z$.