2003 AMC 10A Problem 24

With $5$ red and $4$ blue cards alternating, the stack must be red, blue, red, blue, red, blue, red, blue, red. So both ends are red, and every blue card has a red card on each side.

The cards with the fewest choices control everything, so find those first. Write $R5$ for red card $5$, $B6$ for blue card $6$, and so on.

1. List which blues each red can touch (the red number must divide the blue number):

2. Place $R5$. It can touch only one blue card, but a red card in the middle has two blue neighbors. So $R5$ is at an end, with $B5$ next to it.
3. Place $R4$. Same reason: it is at the other end, with $B4$ next to it.
4. Fill in next to $B5$. Its other neighbor is a red card that divides $5$. $R5$ is used, so it is $R1$.
5. Fill in next to $B4$. Its other neighbor is a red card that divides $4$. $R1$ and $R4$ are used, so it is $R2$.
6. So far:
$$R5,\ B5,\ R1,\ \_,\ \_,\ \_,\ R2,\ B4,\ R4.$$
The cards left are $B3$, $R3$, $B6$, and the middle three spots are blue, red, blue.
7. The blue card next to $R2$ must be a multiple of $2$, so it is $B6$. Then $B3$ goes next to $R1$, and $R3$ goes in the center. Check: $3$ divides both $3$ and $6$. It works.
8. The full stack is $R5, B5, R1, B3, R3, B6, R2, B4, R4$. The middle three cards are $B3, R3, B6$, and $3 + 3 + 6 = 12$.

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

In arrangement puzzles, start with the pieces that have the fewest options. A card with only one allowed neighbor has to be at an end, and each card you place removes choices for the next one, so the whole stack falls into place. Making a table of allowed neighbors first shows you where to start.

Nine cards alternate in color with five red and four blue, so the pattern is $R\,B\,R\,B\,R\,B\,R\,B\,R$: red cards at both ends, and every blue card has two red neighbors that must divide it.

Write $R_k$ and $B_k$ for the red and blue cards numbered $k$. Which blues can each red touch?

• $R_5$: only $B_5$. $\quad R_4$: only $B_4$. $\quad R_3$: $B_3$ or $B_6$.
• $R_2$: $B_4$ or $B_6$. $\quad R_1$: any blue.

$R_5$ has a single possible neighbor, so it must be at an end with $B_5$ beside it; likewise $R_4$ is at the other end with $B_4$ beside it. Reading inward: $B_5$'s other neighbor must divide $5$, so it is $R_1$. $B_4$'s other neighbor must divide $4$ and $R_1$ is used, so it is $R_2$.

The stack so far is $R_5, B_5, R_1, \_, \_, \_, R_2, B_4, R_4$ with $B_3, R_3, B_6$ to place. The blue next to $R_2$ must be a multiple of $2$, so it is $B_6$; then $B_3$ goes beside $R_1$, and $R_3$ sits in the middle (it divides both $3$ and $6$).

Full stack: $R_5, B_5, R_1, B_3, R_3, B_6, R_2, B_4, R_4$. The middle three cards are $B_3, R_3, B_6$, with sum $3 + 3 + 6 = 12$.

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