2025 AMC 10B Problem 21

Write red $=0$, blue $=1$, yellow $=2$, and read colours modulo $3$. The three rules become a single rule:

> every square has a neighbour whose colour is one more than its own.

Name the cells: the centre $M$; the four edge cells $N, E, S, W$; and the four corners, each named by the two edge cells it sits between, so $NE$ lies between $N$ and $E$, and so on. A corner touches only two edge cells; the centre touches only the four edge cells.

Step 1: what a corner can do. Let a corner lie between edge cells of colours $a$ and $b$. Its own colour $k$ needs a neighbour $k+1$, and its only neighbours are those two edge cells, so
$$k = a - 1 \quad\text{or}\quad k = b - 1 .$$
For that corner to be the neighbour required by the edge cell of colour $a$, we would need $k = a+1$. Since $a+1 \ne a-1$, this forces $a+1 = b-1$, that is $b = a-1$. So:

> a corner can serve an edge cell $X$ only when the edge cell on the corner's other side is coloured one less than $X$, and in that case the corner's colour is forced to be $x+1$.

The centre serves an edge cell $X$ exactly when $M = x+1$, and the centre itself needs some edge cell coloured $M+1$.

Step 2: all three colours appear among $N, E, S, W$. Suppose a colour $c$ appears on an edge cell while $c-1$ appears on none. By Step 1 no corner can serve those cells, so the centre must, giving $M = c+1$; then the centre needs an edge cell of colour $M+1 = c+2$.

• If the edge cells use only one colour $c$, then $c-1$ is missing, so $M = c+1$ and a cell of colour $c+2$ is required — there is none.
• If they use exactly two colours $p$ and $q$, then $p-1 = q$ and $q-1 = p$ cannot both hold (they would give $p = p+2$). So some used colour $c$ has $c-1$ missing, forcing $M = c+1$ and requiring an edge cell of colour $c+2 = c-1$, which is missing.

Both are impossible, so the four edge cells carry all three colours: one colour $r$ twice and the other two once each.

Step 3: the centre is forced. There are two shapes.

$r$ on opposite edge cells. Say $N = S = r$; then $\{E, W\} = \{r+1, r+2\}$, and by reflecting the picture take $E = r+1$, $W = r+2$. Both cells flanking $E$ and both flanking $W$ are coloured $r$. A corner serves $E$ only if $r = (r+1)-1$, which is true; it serves $W$ only if $r = (r+2)-1 = r+1$, which is false. So $W$ must be served by the centre: $M = W + 1 = r$. The centre then needs colour $r+1$, supplied by $E$.

$r$ on adjacent edge cells. Say $N = E = r$, with $S$ and $W$ carrying the other two colours. A corner serves $E$ only if one of its flanking cells, $N = r$ or $S$, equals $r-1$, i.e. only if $S = r-1$; likewise a corner serves $N$ only if $W = r-1$. Exactly one of $S, W$ equals $r-1$, so exactly one of $N, E$ must fall to the centre — and either way $M = r+1$. The centre then needs colour $r+2$, which is present.

Step 4: filling in the corners. Take $r = 0$; the other two values of $r$ simply relabel the colours.

Opposite case: $N = S = 0$, $E = 1$, $W = 2$, $M = 0$. The allowed corner colours from Step 1 are
$$NE \in \{2,0\},\quad SE \in \{2,0\},\quad NW \in \{2,1\},\quad SW \in \{2,1\}.$$
- $N = 0$ needs a $1$: not $NE$, not $M = 0$, so $NW = 1$. Similarly $SW = 1$.
- $W = 2$ needs a $0$: the centre supplies it.
- $E = 1$ needs a $2$: so at least one of $NE, SE$ equals $2$.
- $M = 0$ needs a $1$: $E$ supplies it.

Thus $(NE, SE)$ is $(2,2)$, $(2,0)$ or $(0,2)$: three colourings.

Adjacent case: $N = E = 0$, $S = 1$, $W = 2$, $M = 1$. The allowed corner colours are
$$NE \in \{2\},\quad SE \in \{2,0\},\quad SW \in \{0,1\},\quad NW \in \{2,1\}.$$
- $N$ and $E$ are served by the centre, which is $1$.
- $S = 1$ needs a $2$: only $SE$ can be, so $SE = 2$.
- $W = 2$ needs a $0$: only $SW$ can be, so $SW = 0$.
- $NE$ is forced to $2$; $M = 1$ needs a $2$, supplied by $W$.
- $NW$ is free: either $1$ or $2$.

Thus there are two colourings.

Step 5: dividing by the eight symmetries. Fix the repeated colour $r$; there are $3$ choices.

Adjacent case. The eight placements of the pattern (four rotations, then the mirror image $S \leftrightarrow W$) form one orbit, and no non-identity symmetry fixes a placement: such a symmetry would have to fix the two edge cells carrying the unrepeated colours separately, and the only symmetry fixing $S$ other than the identity is the reflection in the line $NS$, which moves $W$. So the $8 \times 2 = 16$ colourings split into $2$ classes.

Opposite case. The four placements form one orbit, and each is fixed by exactly one non-identity symmetry: the reflection in the line $EW$, which swaps $N \leftrightarrow S$, hence $NE \leftrightarrow SE$ and $NW \leftrightarrow SW$. That reflection fixes the completion $(NE,SE) = (2,2)$ and interchanges $(2,0)$ with $(0,2)$, so the $4 \times 3 = 12$ colourings split into $2$ classes.

Each choice of $r$ therefore gives $2 + 2 = 4$ classes, and
$$3 \times 4 = 12 .$$

The answer is $\boxed{\textbf{(C)}\ 12}$.

Naming the colours $0, 1, 2$ removes the temptation to treat red, blue and yellow as three separate cases and exposes the cycle as a single "+1" condition, which is what makes Step 1 possible. The corners are the right place to start because they have the fewest neighbours, so their colours are pinned down to two possibilities before anything else is known; the centre, with four neighbours, is the most flexible cell and is settled last. The same "low degree first" strategy works for any adjacency-constraint colouring on a small board. As a check on the arithmetic, the counts above give $3 \times (16 + 12) = 84$ labelled colourings, which is not a multiple of $8$ — a reminder that symmetric colourings exist and that no single division can replace the orbit count.

• Counting the labelled colourings and dividing by 8, which fails because three of the twelve classes are left fixed by a reflection; 84 is not even divisible by 8.
• Treating the conditions as symmetric, so that a red square next to a blue square is taken to satisfy both squares at once; the cycle red $\to$ blue $\to$ yellow $\to$ red has a direction.
• Forgetting that a corner left "free" at the end still has to satisfy its own condition, which is what restricts it to the two values $a-1$ and $b-1$ in the first place.