Determine the following property For an integer $k \geq 3$, determine all sets $A=\{a1,...,a_k\}$ of  positive integers that have the following property: Whenever h,i,j are distinct then each of $a_h,a_i,a_j$ divides $a_h+a_i+a_j$. 
 A: let $a_1\leq a_2\leq\dots \leq a_k$ be the $k$ positive integers. Then we have that $a_i+a_j$ is a multiple of $a_k$ for $1\leq i<j<k$. This means $a_i\equiv-a_j$ for all $1\leq i<j<k$. 
If $k>3$ this implies $a_1=a_2=a_3\dots=a_{k-1}$.(To see this notice $a_2=a_3\dots =a_{k-1}$, because they are all congruent to $-a_1\bmod a_k$, since $a_k$ is the largest one the equality follows, analogously $a_1=a_2=\dots=a_{k-2}$ because they are all congruent to $-a_{k-1}\bmod a_k$).
So our set consists of an integer $a$, repeated $k-1$ times and an integer $b\geq a$ that appears once. It is easy to see that $b$ must be a multiple of $a$. So $b=am$, and $am$ must divide $a+a+am$, so $m|(m+2)$, meaning $b=a$ or $b=2a$.
If $k=3$ then there are more options, let the numbers be $x\leq y\leq z$. Suppose $x$ and $y$ have already been selected, then we must have that $z$ is a divisor of $x+y$, now, notice $x+y\leq 2y$ (with equality only if $x=y$, So if $x$ is not $x+y$ (but a proper divisor) then $z=y=x$.
In the other case the set is $x,y,x+y$, so now all that remains is to see for which values of $x,y$ we have that $x$ and $y$ divide $2(x+y)$. Clearly this is equivalent to asking $x$ to divide $x+2y$ and $y$ to divide $x+2y$, which again, is equivalent to asking $x$ divides $2y$ and $y$ divides $2x$. So $x=y$ or $2x=y$. 
From here the solutions are:
$\{x,2x,3x\}$
$\{x,x,x\}$
$\{x,x,2x\}$
