three distinct positive integers $a, b, c$ such that the sum of any two is divisible by the third I need to determine three distinct positive integers $a, b, c$ such
that the sum of any two is divisible by the third.
I tried like with out loss of generality let $a<b<c$
As, $a\mid (b+c)$ so $b+c=ak_1$ for some $k_1\in\mathbb{N}$ similarly $$a+b=k_2 c$$ $$a+c=k_3b$$ so adding them I get $k_1+k_2+k_3=2$, could anyone help me to proceed?
 A: $$ak_1-b-c=0 \ \ \ \ \ (1)$$
$$a-bk_2+c=0 \ \ \ \ (2)$$
$$a+b-ck_3=0 \ \ \ \ (3)$$
Using this, for non-trivial solution,
$$\det\begin{pmatrix} k_1 & -1 & -1 \\ 1 & -k_2 &1 \\ 1&1&-k_3\end{pmatrix}=0$$
$$\implies k_1(k_2k_3-1)-1(k_3+1)+1(1-k_2)=0$$
$$\implies k_1+k_2+k_3=k_1\cdot k_2\cdot k_3$$
From the answer by BrianM.Scott of this, we can take $k_1=1,k_2=2,k_3=3$ for any  base $b$ where $b$ is a natural number $>1$
Solving $(1),(2),(3)$ we get $b=2c,a=3c\implies (a,b,c)=(3c,2c,c)$ 
A: You should get $2(a+b+c)=k_1a+k_2b+k_3c $ - $(*)$
Now if $k_1=k_2=k_3=2$ we get $a=b=c$.
Suppose $k_1\le k_2\le k_3$, then we must have $k_3 \gt 2$ and $k_1 \lt 2$ else the two sides of $(*)$ cannot be equal.
Since $k_1$ is a non-zero positive integer less than $2$ is must be $1$.
So we have $a=b+c$, and $b+(b+c)=k_2c$, and $c+(b+c)=k_3b$ so that $$b=\frac {k_2-1}2c, \text{ } c=\frac {k_3-1}2b$$ whence $$(k_2-1)(k_3-1)=4$$ and we have $k_2=k_3=3$ with $b=c$, which is not allowed, or $k_2=2, \text { } k_3=5$.
In this case $c=2b$ and $a=b+c=3b$ gives the family of solutions suggeted in the comments.
A: Since $0+0 <  a + b < c + c = 2c$ and $ c | a + b $, hence $ a + b = c$.
Since $ a < b$, hence $ 2a < a+b =c < 2 b $.
Since $0+ b < a + c < b + 2b $, and $b|a+c$ hence $ a+c = 2b$.
Solving the system of equations,  we get that $b=2a, c=3a$. It's easy to check that $\{a, 2a, 3a\}$ satisfies the conditions.
A: First, note that $0<a<b<c$ and $c|a+b$ implies $a+b=c$.
Since $b|a+c=2a+b$ we have $\beta b=2a+b$ and similarly $\alpha a=a+2b$ where $\alpha,\beta\in\mathbb N$.
Now $a<b$ implies $\beta b<3b$, so $\beta$ must equal $2$ and therefore $2a=b$.
It follows that $c=a+b=3a$.
So the solutions are $(a,2a,3a)$ for any $a\in\mathbb N_+$
A: we are given $a<b<c$, $0<a+b<c+c=2c$ and $c\mid a+b$ so $a+b=c$, now $2a+b=a+c$ and we know $b\mid a+c$ so $b\mid 2a$. Since $2a<2b$ so $b=2a$ and hence $c=a+b=3a$ so $n,2n, 3n$ will work for any $+ve$ integer $n$  
