<s>[INST] Finding the sum of all solutions $2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.
 [/INST] Let us verify for $n=m[2,3]+k$ where $0\le k<[2,3]$
If $k=0\implies 2x+3y=6m\implies 3(2m-y)=2x,$ so $y$  must even $=2z$(say),
So, $x=3(m-z)\implies 0\le z\le m$ so $z$(hence $y$) can assume $m+1$ values in non-negative integers.
If $k=1\implies 2x+3y=6m+1\implies 3(y-2m)+1=2x,$ so $y$  must odd $=2z+1$(say), where $z\ge 0$
So, $x+1=3(m-z)\implies (m-z)\ge 1\implies 0\le z\le m-1$ so $z$(hence $y$) can assume $m$ values in non-negative integers.
$\implies n=6(m+1)+1=6m+7$ will have $m+1$ solutions  in non-negative integers.
So, we observe that $6m,6m+2,6m+3,6m+4,6m+5,6m+7$ will have  $m+1$ solutions in non-negative integers.
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