, 2024

Solution:

Note that $a+b+c=28$ is achieved when $(a, b, c)=(19,6,3)$. To show we cannot do better, first observe we would need $a+c<27$ and $\operatorname{lcm}(a, c) \leq a c \leq 13 \cdot 13=169$, which is only possible when $\operatorname{lcm}(a, c)$ is $7, 17, 57, 77$, or $117$. We do casework on each value:

- $\operatorname{lcm}(a, c)=7$. Then $a$ and $c$ are both $1$ or $7$, so $\operatorname{lcm}(a, b)$ and $\operatorname{lcm}(b, c)$ are both $b$ or $7b$. It is impossible for one of $b$ and $7b$ to end in $4$ and the other to end in $6$.

- $\operatorname{lcm}(a, c)=17$. The same argument as above proves this case is also impossible.

- $\operatorname{lcm}(a, c)=77$. Since the divisors of $77$ only end in $1$ and $7$, the same argument as above rules out this case.

- $\operatorname{lcm}(a, c)=57$. As $a+c<27$, we must have $\{a, c\}=\{3,19\}$. Then $b$ is even and less than $6$, so it's easy to verify there are no solutions here.

- $\operatorname{lcm}(a, c)=117$. As $a+c<27$, we must have $\{a, c\}=\{9,13\}$. Then $b$ is even and less than $6$, so it's easy to verify there are no solutions here.

This rules out all cases, so $28$ is optimal.