Determine three three-digit natural numbers whose decimal representation uses nine different (non-zero) digits such that their product ends with four zeros.
We can check that $625,384,971$.

It is necessary for the product to be divisible by $5^{4}$ and $2^{4}$, which requires choosing 625 (indeed, there cannot be two multiples of 5 among the $3 \mathrm{ni}$, nor any multiple of $5^{4}=625$). We then look for a multiple of 16 with no digits in common with the previous one: 384. We can choose the last number at random.