AMC10 2014 B

Answer (B): The circular arrangement 14352 is bad because the sum $6$ cannot be achieved with consecutive numbers, and the circular arrangement 23154 is bad because the sum $7$ cannot be so achieved. It remains to show that these are the only bad arrangements. Given a circular arrangement, sums $1$ through $5$ can be achieved with a single number, and if the sum $n$ can be achieved, then the sum $15-n$ can be achieved using the complementary subset. Therefore an arrangement is not bad as long as sums $6$ and $7$ can be achieved. Suppose $6$ cannot be achieved. Then $1$ and $5$ cannot be adjacent, so by a suitable rotation and/or reflection, the arrangement is $1bc5e$. Furthermore, $\{b,c\}$ cannot equal $\{2,3\}$ because $1+2+3=6$; similarly $\{b,c\}$ cannot equal $\{2,4\}$. It follows that $e=2$, which then forces the arrangement to be $14352$ in order to avoid consecutive $213$. This arrangement is bad. Next suppose that $7$ cannot be achieved. Then $2$ and $5$ cannot be adjacent, so again without loss of generality the arrangement is $2bc5e$. Reasoning as before, $\{b,c\}$ cannot equal $\{3,4\}$ or $\{1,4\}$, so $e=4$, and then $b=3$ and $c=1$, to avoid consecutive $421$; therefore the arrangement is $23154$, which is also bad. Thus there are only two bad arrangements up to rotation and reflection.