The Problems of Ukrainian Authors

Let in a cell with number $n$ number $a_n$ is written down. We will consider such sequence $(a_n)$: $a_{2k-1}=2k-1$, $k=\overline{1,1004}$, $a_{2k}=2008-2k$, $k=\overline{1,1003}$. We will prove that such placing approaches us. Let $S_n=a_1+a_2+\cdots +a_n$. Then $S_{2k}=2007k$, $k=\overline{1,1003}$, $S_{2k-1}=S_{2k-2}+a_{2k-1}=2009k-2008$, $k=\overline{1,1004}$. And as $S_{2k-1}\equiv -2007k(\mathrm{mod}2008)$, from equality $S_n\equiv S_M(\mathrm{mod}2008)$ follows that $m=n$. Therefore the sequence $(a_n)$ satisfies the condition that for any non-negative integer numbers $i$ and $j$ such that $1\le i<j\le 2007$, the sum of numbers of cells $i,i+1,\dots ,j$ does not divide by $2008$.

Now we will describe a winning strategy for the first player. The first move he marks a cell with number $1004$. Further, if the second player moves in the cell with number $i$, the first player moves into the cell with number $2008-i$. As the sum of numbers with numbers $i$ and $2008-i$ is equal to $2008$ for all $i=\overline{1,2007}$, it is easy to see that the described strategy for the first player is winning.

We will show how from this sequence to receive $999$ more other sequences which also satisfy the statement of the problem. Let $(l,2008)=1$. We put $b_n=l\cdot a_n(\mathrm{mod}2008)$, $n=\overline{1,2007}$. Then the sequence $(b_n)$ obviously satisfies the condition that $\forall i,j\in \mathbb{N}:1\le i<j\le 2007$, the sum of numbers of cells $i,i+1,\dots ,j$ does not divide by $2008$. On the other hand, the strategy of the first player does not change. It is obvious that for different $l=\overline{1,2007}$, we receive different sequences $(b_n)$. Therefore, we receive $\phi (2008)=1000$ sequences.