2023 AMC 10A Problem 20

The center square belongs to all four $2 \times 2$ blocks, so no other square may share its color. Choose the center's color: $4$ ways. The remaining eight squares use only the other $3$ colors.

Label the edge-middle squares $T$ (top), $R$ (right), $B$ (bottom), $L$ (left). Each $2 \times 2$ block contains two of them that are adjacent around the ring ($T$ with $L$, $T$ with $R$, $R$ with $B$, $B$ with $L$), and those two must differ. Opposite squares ($T$ and $B$, or $L$ and $R$) share no block and may match. So we need colorings of a $4$-cycle with $3$ colors where neighbors differ:

• Both opposite pairs equal ($T = B$, $L = R$, different colors): $3 \cdot 2 = 6$.
• Exactly one opposite pair equal: choose which pair ($2$), its color ($3$), then the other two squares take the remaining two colors in either order ($2$): $12$.
• Neither pair equal: impossible, since $T, R, B, L$ would need $3$ colors with $T \ne B$ and $L \ne R$ and all four neighbors distinct, requiring $4$ colors.

That gives $6 + 12 = 18$ ring colorings.

Each corner lies in exactly one block, together with the center and two adjacent ring squares, all three of different colors; the corner must take the fourth color, so it is forced.

Total: $4 \cdot 18 = 72$.

The answer is $\boxed{\textbf{(D)}\ 72}$.

Start with the most constrained cell (the center, in every block), then the next tier (edge squares, each in two blocks), and let the least constrained cells (corners, one block each) be determined last. Reducing the middle tier to "proper colorings of a cycle" is the reusable step; for a $4$-cycle with $k$ colors the count is $(k-1)^4 + (k-1)$, which for $k = 3$ is $18$.

• Treating the edge-middle squares as independent choices, or forgetting the two cases (two opposite pairs equal vs. one pair equal) in the cycle count.
• Counting the corners as free choices, or believing each of the $4$ blocks can be colored in $4!$ ways independently.