1. Each cell of a $68 \times 68$ table is to be colored with one of three colors (red, blue, white). In how many ways can this be done so that every triplet of adjacent cells in each row and each column contains cells of all three colors?

(Josef Tkadlec)
Solution. Once we determine the colors of some two adjacent fields in a row or column of the table, the coloring of all its other fields is uniquely determined by the requirements of the problem. The colors in each row and column thus alternate regularly with a period of 3.

Let's say that $a b c a b c .$. is the coloring of the fields of the first row. Out of the six possible colorings of the fields of any row

abcabc...,
bcaacb...,
cababc...,
acbacb...,
bacacb...,
cbacba...

clearly, only the second and third are suitable for the second row. Once we choose one of them for the second row, the other must be the coloring of the fields of the third row. The fourth row will then be colored the same as the first, the fifth as the second, etc., with a period of 3.

Let's summarize our considerations. For the coloring $a b c a b c .$. of the first row, we have $3 \cdot 2=6$ possibilities (3 for choosing $a$, 2 for choosing $b$). For the coloring of the second row, as we have found, we then have two possibilities. The coloring of the subsequent rows is already uniquely determined. The number of all valid colorings of the entire table is thus equal to $6 \cdot 2=12$.

For a complete solution, award 6 points. For noting that the colors in each row alternate regularly with a period of 3 (even without a detailed explanation), award 2 points and another point for justifying that the number of ways to color one row is thus equal to 6. Award another 2 points for justifying why, after coloring one row, the next row can be colored in two ways, and the final point for concluding that the number of colorings is equal to 12. For merely guessing the correct result, award 1 point. (The same considerations obviously apply to columns as well.)