\begin{tabular}{ | c | c | c | c | c | c | c | c |} \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2\\ \hline \end{tabular}As above, when tilling a $1x3$ trimino, the sum of numbers which are hidden is divisible by 3. Because the sum of all above numbers is also divisible by three, the unit square is on 3-collums. We can find its rows by the same way. The rest is giving an example for each one which is your turn

you can take this one \begin{tabular}{ | c | c | c | c | c | c | c | c |} \hline a & b & c & a & b & c & a & b\\ \hline b & c & a & b & c & a & b & c\\ \hline c & a & {\bf b} & c & a & {\bf b} & c & a\\ \hline a & b & c & a & b & c & a & b\\ \hline b & c & a & b & c & a & b & c\\ \hline c & a & {\bf b} & c & a & {\bf b} & c & a\\ \hline a & b & c & a & b & c & a & b\\ \hline b & c & a & b & c & a & b & c\\ \hline \end{tabular} just by counting we get that the unit square is located at b-cell, after rotating $90$ degrees you get the same but the only overlaping b's are those marked bold. The rest is providing an example which is trivial.

Italy TST 1995