2014 AMC 10B Problem 24

Call a group of numbers that sit next to each other around the circle a run.

Step 1: shrink the list of sums to check.
1. Sums $1$ to $5$ are always possible: each is a single number.
2. Sum $15$ is always possible: take all five.
3. If you take a run, the numbers you did not take also sit next to each other, so they form a run too. The two runs add to $1 + 2 + 3 + 4 + 5 = 15$. For example, if a run sums to $6$, the rest sums to $9$.
4. So sums $8$ to $14$ are possible exactly when $7$ down to $1$ are. The only sums that can fail are $6$ and its partner $9$, and $7$ and its partner $8$.

An arrangement is bad exactly when it cannot make $6$ or cannot make $7$.

Step 2: how can $6$ and $7$ be made? A run of four has sum at least $1 + 2 + 3 + 4 = 10$, so only runs of two or three numbers matter.
- $6$: the pair $1, 5$; the pair $2, 4$; or the three numbers $1, 2, 3$ in a row.
- $7$: the pair $2, 5$; the pair $3, 4$; or the three numbers $1, 2, 4$ in a row.

Step 3: how many arrangements are there? Put $1$ at the top. The other four numbers can go clockwise in $4 \times 3 \times 2 \times 1 = 24$ orders, and flipping the circle over pairs them up, so there are $12$ different arrangements. Rather than check all $12$, ask what a bad one would need.

Step 4: can $6$ fail? For $6$ to fail, $5$ is not next to $1$, and $2$ is not next to $4$, and $1, 2, 3$ are not a run. So $1$'s two neighbors come from $2, 3, 4$.
- Neighbors $2$ and $3$: then $2, 1, 3$ is a run summing to $6$. Good arrangement.
- Neighbors $3$ and $4$: the other two spots hold $2$ and $5$. Either $1, 3, 2, 5, 4$ (run $1,3,2$ makes $6$) or $1, 3, 5, 2, 4$ ($2,4$ are next to each other). Both good.
- Neighbors $2$ and $4$: the other two spots hold $3$ and $5$. Either $1, 2, 3, 5, 4$ (run $1, 2, 3$) or $1, 2, 5, 3, 4$. In the last one, the neighboring pairs ($1\,2$, $2\,5$, $5\,3$, $3\,4$, $4\,1$) sum to $3, 7, 8, 7, 5$, and the runs of three ($1\,2\,5$, $2\,5\,3$, $5\,3\,4$, $3\,4\,1$, $4\,1\,2$) sum to $8, 10, 12, 8, 7$. No $6$. Bad.

Step 5: can $7$ fail? For $7$ to fail, $5$ is not next to $2$, $3$ is not next to $4$, and $1, 2, 4$ are not a run. So $5$'s two neighbors come from $1, 3, 4$.
- Neighbors $3$ and $4$: the other spots hold $1$ and $2$, giving $5, 3, x, y, 4$. With $x = 1$: $3, 1, 2, 4$ puts $1, 2, 4$ in a run. With $x = 2$: the circle $5, 3, 2, 1, 4$ has $2, 1, 4$ as a run. Both good.
- Neighbors $1$ and $3$: the circle is $5, 1, x, y, 3$ with $x, y$ being $2, 4$. If $y = 4$, then $4$ sits next to $3$. So $x = 4$, $y = 2$: the circle $5, 1, 4, 2, 3$ has $1, 4, 2$ in a run. Good.
- Neighbors $1$ and $4$: the circle is $5, 1, x, y, 4$ with $x, y$ being $2, 3$. If $y = 3$, then $3$ sits next to $4$. So $x = 3$, $y = 2$: the circle $5, 1, 3, 2, 4$. Its pairs sum to $6, 4, 5, 6, 9$ and its runs of three ($5\,1\,3$, $1\,3\,2$, $3\,2\,4$, $2\,4\,5$, $4\,5\,1$) sum to $9, 6, 9, 11, 10$. No $7$. Bad.

Step 6: count. The two bad arrangements are $1, 2, 5, 3, 4$ (no $6$) and $1, 3, 2, 4, 5$ (the same circle as $5, 1, 3, 2, 4$; no $7$). They are different, since one can make $6$ and the other cannot.

The answer is $\boxed{\textbf{(B)}\ 2}$.

On a circle, whatever is left after removing a run is also a run, so every sum comes with a partner that adds to $15$. That cuts fifteen checks down to two, sums $6$ and $7$. Then asking "what would have to be true for $6$ to fail?" limits the neighbors of one key number and leaves only a few circles to check by hand. Look for a fact like this that shrinks the work before you start listing.

Just list all $12$ circles. Start at $1$, go clockwise, and to avoid counting a circle and its mirror image twice, keep only the orders where $2$ comes before $5$. For each, look for a run adding to $6$ and a run adding to $7$. Ten of the twelve have both. Only $1, 2, 5, 3, 4$ (no run makes $6$) and $1, 3, 2, 4, 5$ (no run makes $7$) fail, so there are $2$ bad arrangements.