22nd Korean Mathematical Olympiad Final Round

For each coin in the table we give a positive integer one by one from the leftmost one. We show that if the black coin is in the $s=1005$th place, then we can make each coin's black side up after some steps. If we choose $s,s-1,s+1$ one by one, then all five coins having numbers from $s-2$ to $s+2$ become black side up.

Assume that all coins with numbers $$s-k,s-(k-1),\dots ,s-1,s,s+1,\dots ,s+(k-1),s+k$$ become black side up. If $k=2\ell$ then we proceed the steps one by one with black coins having the following numbers $$s-2\ell ,s+2\ell ,s-2(\ell -1),s+2(\ell -1),s-2,s+2,s.$$ and if $k=2\ell -1$ then $$s-(2\ell -1),s+(2\ell -1),s-(2\ell -3),s+(2\ell -3),s-1,s+1.$$ so that we can make all coins with numbers from $s-k-1$ to $s+k+1$ become black side up.

Now assume that all coins whose black sides are up have numbers $T_1,T_2,\dots ,T_k$ in some state. We define $$M=\sum _{i=1}^k(-1{)}^{i+1}{T}_i=T_1-T_2+T_3-\cdots +(-1{)}^{k-1}{T}_k.$$ Assume that we choose the coin having a number $T_j$ to proceed the step. First consider the case when $T_j\ne 1,2009$. If $T_{j-1}\ne T_j-1$ and $T_{j+1}\ne T_j+1$ then the value $M$ after the procedure becomes $$\sum _{i=1}^{j-1}(-1{)}^{i+1}{T}_i+(-1{)}^{j+1}(T_j-1)+(-1{)}^{j+2}{T}_j+(-1{)}^{j+3}(T_j+1)+\sum _{i=j+1}^k(-1{)}^{i+1}{T}_i.$$ which equals to $M$. For the remaining one may easily check that $M$ is an invariant in each step. If $j=1$ then $M$ is changed by $-M$ whether or not $T_2=2$. Finally if $j=2009$, then one may easily check that $M$ is changed by $M+(-1{)}^{k+1}2010$. Therefore the remainder $r$ or $2010-r$ of $M$ divided by $2010$ is invariant under the procedure.

If every coin becomes black side up then $M=1-2+...+2007-2008+2009=1005$. Therefore only when the coin whose black side is up is in the $1005$th place, we can make every coin's black side up. $\square$