Find all sets $a_1,…,a_k$ of positive integers such that the sum of any triplet is divisible by each member of the triplet.

Let $k\geq3$ be a fixed integer. Find all sets $a_1,...,a_k$ of positive integers such that the sum of any triplet is divisible by each member of the triplet.

I have no idea of how to start attacking this problem. Could someone give me a hint?

Thank you.

Consider a qualifying triplet $\{a,b,c\}$ with $a<b<c$. Then clearly $a+b<2c$ and thus in order for $c \mid a{+}b{+}c$ we need $a+b=c$.

If we add a fourth optional value, $d>c$, we know that $a+b<d$ and so this triplet will fail. So only a set with $k=3$ can possibly work.

Now we know $a+b+c=2c$ and $b<c<2b$ so $b \mid a{+}b{+}c \implies 3b=2c$ and thus $6a=2c$. So all such sets are of the form $\{a,2a,3a\}$.

First notice that if you have a triplet $(a,b,c)$, then

$$a|b+c$$ $$b|a+c$$ $$c|a+b$$

If the triplet is ordered increasingly, then, you may show that

$$c=a+b$$

With the new information, you get the triplet $(a,b,a+b)$. Therefore

$$a|2b+a$$ $$b|2a+c$$

which means that we need to have

$$a|2b$$ $$b|2a$$

Having $a<b$ and $b|2a$, show that $b=2a$.

The second part is to show that you cannot have more members in the set, for if you add a new integer $d$, to be the greatest in the set, $(a,b,d)$ will not make a triplet.