
3. Find all possible triples $(a, b, c)$ of positive integers with the following properties:

- $\operatorname{gcd}(a, b)=\operatorname{gcd}(a, c)=\operatorname{gcd}(b, c)=1$;
- $a$ is a divisor of $a+b+c$;
- $b$ is a divisor of $a+b+c$;
- $c$ is a divisor of $a+b+c$.

(Here $\operatorname{gcd}(x, y)$ is the greatest common divisor of $x$ and $y$.)


3. The problem is symmetric in $a, b$, and $c$. That is, if we consistently swap $a$ and $b$, or $a$ and $c$, or $b$ and $c$, then the conditions on $(a, b, c)$ do not change. We will therefore consider solutions for which $a \leqslant b \leqslant c$. The remaining solutions are then found by permuting the values of $a, b$, and $c$.

Since $a$ and $b$ are positive, we see that $a+b+c>c$. Since $c$ is largest among the three numbers, we also have $a+b+c \leqslant 3 c$. Since we are given that $a+b+c$ is a multiple of $c$, we are left with two possibilities: $a+b+c=2 c$ or $a+b+c=3 c$. We consider both cases separately. If $a+b+c=3 c$, then $a, b$, and $c$ must all be equal, because otherwise, the fact that $a, b \leqslant c$ implies that $a+b+c0$, we have $2 a+2 b>2 b$. Since $b \geqslant a$, we also have $2 a+2 b \leqslant 4 b$. Therefore, since $2 a+2 b$ must be a multiple of $b$, there are only two possibilities: $2 a+2 b=3 b$ or $2 a+2 b=4 b$. Again, we consider these cases separately.

If $2 a+2 b=3 b$, then $b=2 a$. Similarly to the first case, we find that $a=\operatorname{gcd}(a, 2 a)=\operatorname{gcd}(a, b)=1$. Therefore, $b=2$ and $c=a+b=3$. The resulting triple $(a, b, c)=(1,2,3)$ is indeed a solution, since $\operatorname{gcd}(1,2)=\operatorname{gcd}(1,3)=\operatorname{gcd}(2,3)=1$ and $1+2+3=6$ is divisible by 1,2 , and 3 .

If $2 a+2 b=4 b$, then $a=b$. Again we see that $a=\operatorname{gcd}(a, a)=\operatorname{gcd}(a, b)=1$. From $b=a=1$ it follows that $c=a+b=2$ and hence $(a, b, c)=(1,1,2)$. This is indeed a solution since $\operatorname{gcd}(1,1)=\operatorname{gcd}(1,2)=1$ and $1+1+2=4$ is divisible by 1 and 2 .

We conclude that the solutions for which $a \leqslant b \leqslant c$ holds are: $(a, b, c)=(1,1,1),(a, b, c)=$ $(1,1,2)$, and $(a, b, c)=(1,2,3)$. Permuting the values of $a, b$ and $c$, we obtain a total of ten solutions $(a, b, c)$ :

$$
(1,1,1), \quad(1,1,2),(1,2,1),(2,1,1), \quad(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)
$$
