China Southeastern Mathematical Olympiad

Let $n=\overline{d_1{d}_2{d}_3{d}_4}=1000d_1+100d_2+10d_3+d_4$, $d_1,d_2,d_3,d_4\in [0,1,2,\dots ,9]$, and $S(n)=d_1+d_2+d_3+d_4$. Match the non-negative multiples of 6 less than 2000 into 167 pairs $(x,y)$, $x+y=1998$, such that $(0,1998),(6,1992),(12,1986),\dots ,(996,1002).$ For each pair $(x,y)$, let $x=\overline{a_1{a}_2{a}_3{a}_4}$, $y=\overline{b_1{b}_2{b}_3{b}_4}$, then $$1000(a_1+b_1)+100(a_2+b_2)+10(a_3+b_3)+(a_4+b_4)=x+y=1998.$$ Since $x,y$ are even, $a_4,b_4\le 8$. So $a_4+b_4\le 16<18$. Thus, $a_4+b_4=8$. Since $a_3+b_3\le 18<19$, $a_3+b_3=9$. Similarly, we can obtain $a_2+b_2=9$ and $a_1+b_1=1$. Thus, $$\begin{array}{rl}S(x)+S(y)& =(a_1+b_1)+(a_2+b_2)+(a_3+b_3)+(a_4+b_4)\\ & =1+9+9+8=27.\end{array}$$ Consequently, there is only one of $S(x)$ and $S(y)$ that is the multiple of 6. (This is because that $x$, $y$ are all multiples of 3, so are $S(x)$ and $S(y)$.) That is, there is only one of $x$ and $y$ which is a six match number. Therefore, there are 167 six match numbers less than 2000, and there is just one six match number between 2000 and 2011. Therefore, the answer is $167+1=168$.