Winter Mathematical Competition

We denote the rows by $i=1,\dots ,n$ and the columns by $j=1,\dots ,n$ and let $a(i;j)$ be the number written in the square $(i;j)$.

a) No! Consider the table $A$ where the squares $(i;j)$ are mined if and only if $i\equiv j\equiv 1(\mathrm{mod}3)$ and the table $B$ where the squares $(i;j)$ are mined if and only if $i\equiv j\equiv 2(\mathrm{mod}3)$. Then all numbers written in the squares of $A$ and $B$ are equal to $1$ and therefore we can not determine which squares are mined.

b) Yes! We first determine the mined squares in the third row. It is easy to see that the number $b(j)$ of the mined squares amongst $(3;j-1)$, $(3;j)$, $(3;j+1)$ is equal to $a(2;j)-a(1;j)$. We now compare $b(1)$ and $b(2)$ and determine if the square $(3;3)$ is mined. Next we compare $b(4)$ and $b(5)$ and decide if $(3;6)$ is mined. The same argument shows which squares amongst $(3;9)$, $(3;12)$, ..., $(3;2007)$ are mined.

We now compare $b(2007)$ and $b(2006)$ to decide if the square $(3;2005)$ is mined, then compare $b(2004)$ and $b(2003)$ to see if $(3;2002)$ is mined, etc. We finally know which squares amongst $(3;1999)$, $(3;1996)$, ..., $(3;1)$ are mined. Now $b(1)$ shows if $(3;2)$ is mined, then we compare $b(3)$ and $b(4)$ to determine if $(3;5)$ is mined, etc., and find which squares amongst $(3;8)$, $(3;11)$, ..., $(3;2006)$ are mined. Hence we know row $3$.

Analogously we can determine the mined squares in the rows $6$, $9$, $12$, $15$, ..., $2007$. Using similar argument we can determine the mined squares in the rows $2005$, $2002$, $1999$, ..., $4$, $1$ and the rows $2$, $5$, $8$, ..., $2006$.