3. Determine the number of all triples of distinct three-digit natural numbers, the sum of which is divisible by each of the three addends.
SOLUTION. Let $x, y, z$ be a triplet of distinct natural numbers such that each of them divides their sum $x+y+z$, meaning $x$ divides $y+z$, $y$ divides $x+z$, and $z$ divides $x+y$. Without loss of generality, we can assume $x<y<z$. Therefore, $x+y=kz$ for some natural number $k$. Since $x+y<2z$, it must be that $k=1$, so $x+y=z$. Further, $y$ divides $x+z=2x+y<3y$, so $2x+y=2y$, which implies $y=2x$. The three natural numbers with the given properties thus have the form $x, y=2x, z=3x$, where $x$ is a natural number. Since they must be three-digit numbers, we have $x \geq 100, 3x \leq 999$, so $100 \leq x \leq 333$. The number of such triplets is $333-99=234$.

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