Show that there exists $i\in \lbrace 1, 2, 3 \rbrace $ s.t. there exists $a, b\in A_i $ s.t. $a+b\in B $. Let $A=\lbrace 1, 2, 3,..., 2019\rbrace= A_1\cup A_2\cup A_3$, where $A_1\cap A_2=A_2\cap A_3= A_1\cap A_3=\emptyset $ and $B=\lbrace 672, 1008, 1344, 1680, 2016\rbrace $. 
Show that there exists $i\in \lbrace 1, 2, 3 \rbrace $ s.t. there exists $a, b\in A_i , a\neq b$ s.t. $a+b\in B $. 
I consider $A_1=\lbrace x_1, x_2, ..., x_k \rbrace$. Then $2018-x_1, 2018-x_2,..., 2018-x_k\notin A_1$.But is not enough.
 A: Letting $b=168$, we have $B=\{4b,6b,8b,10b,12b\}$, while $2019=12b+3$. Thus,
$\{b,2b,3b,4b,5b,6b,7b,8b,9b,10b,11b,12b\}\subseteq A$, and restricting to these multiples of
$b$ only, it suffices to show that for any partition
$\{1,2,3,4,5,6,7,8,9,10,11,12\}=A_1\cup A_2\cup A_3$, there are $i\in\{1,2,3\}$ and $a,b\in A_i,\ a\ne b$, such that $a+b\in\{4,6,8,10,12\}$. 
This can be done by a simple case analysis. 


*

*Suppose WLOG that $1\in A_1$. 

*If also $3\in A_1$, then we can take $a=1$ and $b=3$; suppose thus
that $3\in A_2$. 

*If $5\in A_1$, we take $a=1$ and $b=5$; if
$5\in A_2$, we take $a=3$ and $b=5$. Suppose thus that $5\in A_3$. 

*Now if $7\in A_1$, then we take $a=1$ and $b=7$; if $7\in A_2$, we take
$a=3$ and $b=7$; finally, if $7\in A_3$, we take $a=5$ and $b=7$.


Notice, that in this solution we have only used the fact that $\{b,3b,5b,7b\}=\{168,504,840,1176\}\subset A$.
A: Notice that any two of {840,504,168,1176} are not allowed to be in the same $A_i$, then there is a contradiction.
