Prove there is such a triple There are $3n$ colored numbers from $1$ to $3n$ such that each color is represented exactly by $n$ numbers. Prove there is 3-colored triple $a,b,c$ such as $$a+b=c$$
I started by assuming various values for minimum each of color, but didnt come to any conclusions(
 A: Note that this problem was initially asked and solved by Alekseev and Savchev, in the Kvant journal, 4:23, problem M1040.
Let $A,B,C$ be the three monochromatic subsets of $[3n]$. Without loss of generality let $1,\ldots,k-1$ be in $A$ (i.e. the first $k-1$ integers are in $A$, with $k-1\geq1$), and let $k\in B$.
We call three numbers a good triple if they satisfy your condition. Suppose that there are no good triple. 
Let $a\in C$ be any number. Note that $a-1\not\in B$ as otherwise $(1,a-1,a)$ would be a good triple. 
Suppose that $a-1\in C$, and consider 


*

*The integer $a-k$. If it is in $A$ then $(a-k,k,a)$ would be a good triple. If it is in $B$, then $(k-1,a-k,a-1)$  would be a good triple. Therefore $a-k \in C$.

*The integer $a-k-1$. If it is in $A$ then $(a-k-1,k,a-1)$ would be a good triple. If it is in $B$, then $(1,a-k-1,a-k)$  would be a good triple. Therefore $a-k-1 \in C$.

*The integer $a-2k$. If it is in $A$ then $(a-2k,k,a-k)$ would be a good triple. If it is in $B$, then $(k-1,a-2k,a-k-1)$  would be a good triple. Therefore $a-k \in C$.

*...


Repeating this argument we conclude that all integers of the form $a-ik$ and $a-ik-1$ ($i=0,1,\ldots$) are in $C$. But note that because $a>k$, there exist some $i$ such that $ik\leq a \leq i(k+1)$ and therefore such that $1\leq a-ik \leq k$. And we know that this number is either in $B$ (if it is $k$) or in $A$ (if is it $<k$). Hence a contradiction, and $$a-1\in A$$
Therefore we proved that $\forall a\in C, \ a-1 \in A$. But note that $k-1\in A$ while $k\in B$ therefore there is at least one element $a\in A$ for which $a+1\not\in C$. Therefore $\vert A\vert > \vert C\vert$, a contradiction, and there must be at least one good triple.
