Number Divides Pair Sum

If $a \lt b \lt c$ then since $c|a + b$ and $(a + b) \lt 2c$ we must have $a + b = c$ .

This means that $a|b + c \Longrightarrow a|2b + a \Longrightarrow a|2b$

and that $b|c + a \Longrightarrow b|2a + b \Longrightarrow b|2a$ .

This means that (i) $a|2$ or $a|b$ , and (ii) $b|2$ or $b|a$ . But we require that $a,b$ are coprime and assumed that $a \lt b$ , which implies that $a = 1, b = 2$ , and thus $c = 3$ . This gives us a value of $abc = 6$ .

Now we can't have $b = c$ unless $a = b = c = 1$ , for otherwise $c$ would not divide $a + b$ . Also, if $a = b \lt c$ then for $a,b$ to be coprime we would require that $a = b = 1$ , implying that $c = 2$ in order to have $c|a + b$ with $c \gt b = 1$ .

Thus the maximum possible value of $abc$ is $\boxed{6}$ .

If $a|b+c,b|c+a$ and $c|a+b$ then: $a|a+b+c$ $b|a+b+c$ $c|a+b+c$ Therefore: $abcq=a+b+c$ For some $q\in\mathbb{N}$ . The maximum value for $abc$ occurs when $q=1$ , hence: $abc=a+b+c$ Assume that $a<b<c$ , therefore $a+b+c<3c$ . Thus: $abc<3c$ $ab<3$ The only possible values for $(a,b)$ are $(0,1),(0,2)$ and $(1,2)$ . Hence, the only possible solution is $(a,b,c)=(1,2,3)$ $\Rightarrow abc=\boxed{6}$

I just guessed a b and c were 1, 2, and 3, and then multiplied them together and got the answer right, hahaha