Problem 2.5 Determine the natural numbers $a, b, c$ with the property that $a+b+c=a b c$.

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Solution We observe that if one of the numbers is 0, then all the numbers are equal to 0.

$1 \mathrm{p}$

If $a b c \neq 0$, then the relation can be written as $\frac{1}{b c}+\frac{1}{a c}+\frac{1}{a b}=1$. Since the relation is symmetric in $a, b, c$, we can assume $a \leq b \leq c$, from which $a b \leq a c \leq b c$.

If $a b>3$, then $\frac{1}{b c}+\frac{1}{a c}+\frac{1}{a b}<1 . \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .2 p$

If $a b=2$, then $a=1$ and $b=2$, from which we get $\frac{1}{2 c}+\frac{1}{c}=\frac{1}{2}$ and then $c=3$.

If $a b=3$, then $a=1$ and $b=3$, from which we get $\frac{1}{3 c}+\frac{1}{c}=\frac{2}{3}$ and then $c=2$, which does not fit because we assumed $b<c$. ............ 4 p