National Olympiad Final Round

As $\text{gcd}(a,b)=2$, $\text{gcd}(b,c)=3$ and $\text{gcd}(c,a)=5$, the number $a$ is divisible by both $2$ and $5$, the number $b$ is divisible by both $2$ and $3$, and the number $c$ is divisible by both $3$ and $5$. Hence $a$ is divisible by $10$, $b$ is divisible by $6$ and $c$ is divisible by $15$. As $61$ gives remainder $1$ when divided by each of $2$, $3$ and $5$, the numbers $a,b$ and $c$ must give remainder $1$ when divided by $3$, $5$ and $2$, respectively. Since $a,b,c\le 61$, the possibilities are $a=10$ or $a=40$, $b=6$ or $b=36$, and $c=15$ or $c=45$. The sum $61$ appears in three cases: $a=10,b=6,c=45$; $a=10,b=36,c=15$; $a=40,b=6,c=15$. A straightforward check shows that the conditions $\text{gcd}(a,b)=2$, $\text{gcd}(b,c)=3$ and $\text{gcd}(c,a)=5$ are also met in all these cases.