43. Given that $a, b, c$ are positive integers satisfying
$$
a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+\operatorname{gcd}(c, a)+120 \text {, }
$$
determine the maximum possible value of $a$.
Solution: 240 . Notice that $(a, b, c)=(240,120,120)$ achieves a value of 240 . To see that this is maximal, first suppose that $a>b$. Notice that $a+b+c=\operatorname{gcd}(a, b)+\operatorname{gcd}(b, c)+$ $\operatorname{gcd}(c, a)+120 \leq \operatorname{gcd}(a, b)+b+c+120$, or $a \leq \operatorname{gcd}(a, b)+120$. However, $\operatorname{gcd}(a, b)$ is a proper divisor of $a$, so $a \geq 2 \cdot \operatorname{gcd}(a, b)$. Thus, $a-120 \leq \operatorname{gcd}(a, b) \leq a / 2$, yielding $a \leq 240$. Now, if instead $a \leq b$, then either $b>c$ and the same logic shows that $b \leq 240 \Rightarrow a \leq 240$, or $b \leq c, c>a$ (since $a, b, c$ cannot all be equal) and then $c \leq 240 \Rightarrow a \leq b \leq c \leq 240$.