Iranian Mathematical Olympiad

Answer. Yes.

Imagine a $32\times 5$ table $T$ such that the set of its rows is equal to the set of all 5-tuples $(b_0,b_1,\dots ,b_4)$ in which $\forall 0\le i\le 4:b_i\in \{0,1\}$.

Now consider a $32\times 16$ table $R$. Let $(c_i)$ be the $i$-th column and the $j$-th column $(c_j)$ is "consecutive" if and only if $i-j\equiv \pm 1(\mathrm{mod}16)$.

Show the square located in the row $r_i$ and the column $c_j$ with the sign $(i,j)$. Also all the tables made of the first, second and third five columns of $R$ as $T_1,T_2$ and $T_3$ respectively.



Then we will fill the table using the following method: Put a copy of table $T$ in the squares of table $T_1$ (which means that the values of the corresponding squares in $T$ and $T_1$ are the same.) And do the same for table $T_2$ and $T_3$. Then the only empty squares of $R$ are the squares in $c_{16}$.

If we put ${\omega }_{i,16}\equiv {\sum }_{j=1}^5{\omega }_{i,j}(\mathrm{mod}2)$ (1), in which ${\omega }_{i,j}$ means the value of $(i,j)$, then there will be an answer for the problem if we stick $R$ around a cylinder.

Proof. Assuming $S$ as a set of 5 "consecutive" columns in $R$, there can be two conditions:

i) $c_{16}\notin S$: Then we knew that $S$ contains 5 different columns in $S$ is "nice".

ii) $c_{16}\in S$: The other 4 columns in $S$ (except for $c_{16}$,) are copies of 4 different columns in $T$, i.e. assume that they are similar to $c_1,c_2,c_3$ and $c_4$ (2). If there exist two rows $r_i$ and $r_j$ in $S$ ($i\ne j$) with equal values, then using (2) we will have: $\forall 1\le k\le 4:{\omega }_{i,k}={\omega }_{j,k}\Rightarrow$ According to the property that $T$ is nice, ${\omega }_{i,5}\ne {\omega }_{j,5}\Rightarrow s_i\ne s_j$, in which $s_i$ is the sum of $i$-th row in $T$. But $r_i$ and $r_j$ are similar so ${\omega }_{i,16}={\omega }_{j,16}\stackrel{(1)}{⟹}{s}_i=s_j$ which is a contradiction. ■