894. Find three natural numbers whose sum equals their product. List all solutions.
$\triangle$ Let the required numbers be denoted by $x, y$, and $z$. We obtain the equation:

$$
x+y+z=x y z
$$

For natural numbers $x, y$, and $z$, the right-hand side of this equation is generally greater than the left-hand side; therefore, the unknown numbers should be small. We will impose restrictions on these numbers in the form of inequalities.

Let

$$
x \leqslant y \leqslant z
$$

Such restrictions on the unknowns are usually imposed in cases where the equation is symmetric with respect to the unknowns. This does not violate the generality of the reasoning, as the restriction can be removed at the end of the solution.

We get:

$$
x y z=x+y+z \leqslant z+z+z=3 z, \quad x y z \leqslant 3 z
$$

from which

$$
x y \leqslant 3 \text{. }
$$

We will consider several cases related to the last inequality. In doing so, we need to remember that $x \leqslant y$.

1) Let $x=y=1$.

Substitute these values of $x$ and $y$ into the original equation.

$$
1+1+z=z, \quad 2+z=z
$$

The last equation has no solutions.

190

2) Let $x=1, y=2$.

Then

$$
1+2+z=2 z, \quad z=3
$$

3) Let $x=1, y=3$.

We get:

$$
1+3+z=3 z, \quad z=2
$$

But for this case, this is impossible, since it turned out that $z<y$, while it should be $z \geqslant y$.

Obviously, these exhaust all the cases.

Answer: 1, 2, and 3.