2003 AMC 10A Problem 24

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 $\boxed{\textbf{(E)}\ 12}$.

Arrangement puzzles with divisibility rules are solved by attacking the tightest constraints: cards with only one legal neighbor are pinned to the ends, and each placement removes options for the next. Listing the allowed neighbors for every card up front makes the forced chain visible.

• Starting from red 1 (which fits anywhere) rather than from the most restricted cards, and getting lost in possibilities.
• Forgetting that the alternating pattern forces red cards at both ends, and trying to put a blue card at an end.