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The answer is 2. To see this, observe that we can divide the $2 \times 3$ grid into 2 boards, $2 \times 3$ each. In each of them, we can mark at most 2 squares. Because if we could mark 3 squares and if one of the squares in the central column of this $2 \times 3$ grid were not marked, that square would be adjacent to 3 marked squares, which is absurd. If both squares in the central column of the $2 \times 3$ grid are marked, then one of the four corner squares is marked, and thus the square immediately above (or below) it would have three adjacent marked squares, which is absurd. Therefore, for each $2 \times 3$ grid, we have at most 2 marked squares, and the number of marked squares is less than or equal to $2^n$. An example with exactly $2^n$ marked squares is shown below.

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Act 1: Determination of Possible Tile Placements For each of the $m \times n$ boards with 2 marked squares, we number how many marked squares each of the $n$ columns has. As we saw in the previous section, each set of 3 consecutive columns must be numbered with numbers that sum up to at most 2. Therefore, considering the following types of $2 \times 3$ boards based on the numbers of their columns: Note that if we fill the $2 \times 3$ grid with $2 \times 3$ boards (whose possibilities are described by the types shown above), we can place a $2 \times 3$ of type $\alpha$ and immediately after it, a $2 \times 3$ of type $\beta$ if and only if type $\alpha$ is to the left of type $\beta$ in the diagram below, or $\alpha = \beta$: A simple way to see this is to consider whether the sum of three consecutive numbers is at most 2 when comparing each pair. Also, note that we cannot place two $2 \times 3$ boards of types $C_1$ and $C_2$ side by side. Act 2: Calculation of the Number of Ways From here, there are essentially two methods to perform this calculation. Method 1: Direct Calculation by Summation and Ball-Trace If on a $2 \times 3$ board we have $n$ blocks of type $\alpha_1$, $m$ blocks of type $\alpha_2$, $k$ blocks of type $\beta_1$ or $\beta_2$, $l$ blocks of type $\gamma$, and $s$ blocks of type $\delta$ forming the $2 \times 3$ board, then the order in which the types appear is uniquely determined, and thus, the total number of $2 \times 3$ boards formed by these types of blocks is equal to: 1)We count the cases where the types that appear are of type $\alpha_1$. There are 4 ways to choose each $2 \times 3$ of types $\alpha_1$ and $\alpha_2$, and one way to choose the others. 2) We count the cases where the types that appear are of type $\alpha_2$. There are 4 ways to choose each $2 \times 3$ of types $\alpha_1$, $\alpha_2$, and $\delta$, and one way to choose the others. Hence, the total number of ways is Where each summation ranges over all quintuples $(\alpha_1, \alpha_2, \beta_1, \beta_2, \gamma)$ of non-negative integers such that $\alpha_1 + \alpha_2 + \beta_1 + \beta_2 + \gamma = \delta$. Fixing $\alpha_1 + \alpha_2 = \delta$ in the first summation (which results in $(\delta - \alpha_1 + 2^2)$ possibilities for $\alpha_1, \alpha_2$ with $\alpha_1 + \alpha_2 + \beta_1 = \delta - \alpha_1$ and $(\alpha_1 + 1)$ possibilities for $\alpha_2$), and also fixing $\alpha_1 + \alpha_2 + \beta_1 = \delta$ in the second summation (which results in $(\delta - \alpha_2 + 1)$ possibilities for $\alpha_1, \alpha_2$ with $\alpha_1 + \alpha_2 = \delta - \alpha_2$ and $((\alpha_2 + 2) - (\alpha_2 + 1)) = (\alpha_2 + 1)$ possibilities for $\alpha_2, \beta_1, \beta_2$ since the case $\alpha_2 = 0$ has already been counted in (1)), we have:

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