Find the number of solutions to the equation $$100x+10y+z=5xyz$$ where $x,y,z \in \mathbb{Z}$ and $(1 \le x \le 9)$, $(0 \le y \le 9)$, $(0 \le z \le 9)$.
I have found one solution using a brute-force computer program, but I've been wondering if it's possible to find the number of solutions without checking all possible triplets.
If you do arithmetic modulo 5, then all the multiple-of-5 terms ($100x$, $10y$, and $5xyz$) cancel out and you're left with just $z = 0$ (mod 5). IOW, $z$ is a multiple of 5. There are two cases to deal with:
$$z = 0 \implies 100x+10y=0$$ $$z = 5 \implies 100x+10y+5=25xy$$
For the first case, $100x+10y=0$ is equivalent to $y = -10x$. So $y$ must be a multiple of 10, and the only in-range one is $y = 0$. But this gives us $x = 0$, which contradicts the given constraint $1 \le x \le 9$. So any solutions with $z = 0$ are ruled out.
For the $z = 5$ case, solving for $y$ in terms of $x$ gives:
$$y = \frac{20x+1}{5x-2}$$
Or equivalently,
$$y = 4 + \frac{9}{5x-2}$$
So in order for $y$ to be an integer, we must have $5x - 2$ be a divisor of 9. IOW, one of the following:
Only one of these results in an integer: $x = 1$, from which $y = \frac{21}{3} = 7$. Therefore, the only valid all-integer solution is:
$$\boxed{x = 1, y = 7, z = 5}$$