9. Marina wants to color each cell of a $5 \times 5$ table in red or blue, in such a way that each cell shares at most one side with a cell of the same color. How many possible colorings of the table are there?
(A) 30
(B) 45
(C) 46
(D) 50
(E) 90
9. The answer is (A). First, observe that once a row and a column of the table are colored respecting the given conditions, then there is at most one way to complete the coloring while respecting the conditions of the problem: indeed, given a $2 \times 2$ table with three cells colored using both colors, there is a unique way to color the fourth cell while maintaining the conditions (this can be seen by listing the few possible cases). Note also that a $2 \times 2$ table with three cells of the same color contradicts the conditions of the problem.

Therefore, given a coloring on a row and a column of our $5 \times 5$ table, the only way to expand this coloring to the entire table is to look at (one at a time) the $2 \times 2$ subtables of which 3 cells are already colored (necessarily using both colors) and coloring the remaining fourth cell.

Observe that once the first row is colored such that there are no three consecutive cells of the same color, but there are at least two adjacent cells of the same color (which can be done in 14 possible ways), there is a unique way to expand the coloring to the entire table: indeed, if there are two adjacent blue cells, the two immediately below must be red, those further below must be blue, and so on; thus the coloring of at least one column is determined and, as stated above, there is at most one coloring of the entire table that extends the initial coloring. Furthermore, the coloring obtained by repeating the coloring of the first row on the third and fifth rows, and placing the opposite coloring on the second and fourth rows, is an admissible coloring. Note that in this case, the coloring of the first column is obtained by alternating colors. Therefore, there are 14 colorings of this type.

In a completely analogous manner, a coloring of the first column with at least two adjacent cells of the same color (which can always be done in 14 distinct ways) determines a coloring of the entire table. In this case, the first row will have alternating colors and therefore cannot coincide with any of the colorings counted above.

Even in the case where both the first row and the first column are colored with alternating colors, there is a unique coloring possible for the entire table, given by the checkerboard coloring. Therefore, there are 2 colorings that fall into this case.

Since we have exhausted all possible colorings of the first row and the first column, there are no other colorings beyond those listed, and thus the total number of colorings is $14+14+2=30$.