2011 AMC 10B Problem 21

Picture the four numbers on a number line: $z$, then $y$, then $x$, then $w$. Think about the three gaps between neighbors, because every difference is made of them.

1. Name the gaps. Let $a = w - x$, $b = x - y$, $c = y - z$.
2. List all six differences using the gaps.
- Neighbors: $a$, $b$, $c$.
- Two steps apart: $w - y = a + b$ and $x - z = b + c$.
- End to end: $w - z = a + b + c$.
3. The biggest difference is end to end. So $a + b + c = 9$.
4. Which three differences are the gaps? They are three of $1, 3, 4, 5, 6$ that add to $9$.
- If $6$ were a gap, the other two would add to $3$. No two numbers in the list do that.
- If $4$ were a gap, the other two would add to $5$. No two of $1, 3, 5, 6$ do that.
- That leaves $1 + 3 + 5 = 9$. So the gaps are $1, 3, 5$ in some order, and the two-step sums must be the other two numbers, $4$ and $6$.
5. Which gap is in the middle? The two-step sums are $a + b$ and $b + c$, which both use the middle gap $b$.
- Middle $5$: sums $5 + 1 = 6$ and $5 + 3 = 8$. But $8$ is not in the list. No.
- Middle $3$: sums $3 + 1 = 4$ and $3 + 5 = 8$. No.
- Middle $1$: sums $1 + 3 = 4$ and $1 + 5 = 6$. Yes.
So $b = 1$, and the outside gaps are $3$ and $5$ in either order.
6. Case 1: gaps $3, 1, 5$ from the top. The numbers are $z + 9$, $z + 6$, $z + 5$, $z$. Their sum is $4z + 20 = 44$, so $z = 6$. The list is $15, 12, 11, 6$, and $w = 15$.
7. Case 2: gaps $5, 1, 3$ from the top. The numbers are $z + 9$, $z + 4$, $z + 3$, $z$. Their sum is $4z + 16 = 44$, so $z = 7$. The list is $16, 11, 10, 7$, and $w = 16$.
8. Check Case 1: $15 - 12 = 3$, $12 - 11 = 1$, $11 - 6 = 5$, $15 - 11 = 4$, $12 - 6 = 6$, $15 - 6 = 9$. All six match. Case 2 checks the same way.
9. The sum of the possible values of $w$ is $15 + 16 = 31$.

The answer is $\boxed{\textbf{(B)}\ 31}$.

For numbers in order, every difference is a sum of neighboring gaps, and the largest difference is the whole spread. Starting from the largest difference shrinks the search to picking three gaps with a fixed total. When a question asks for "the sum of the possible values," expect more than one answer and look for all of them.

Write the consecutive gaps as $a = w - x$, $b = x - y$, $c = y - z$. The six pairwise differences are then $a, b, c, a + b, b + c$, and $a + b + c$. The biggest of the six is $w - z = a + b + c$, so $a + b + c = 9$.

The three gaps are three of the numbers $1, 3, 4, 5, 6$ adding to $9$. The only such triple is $\{1, 3, 5\}$ (for instance $4 + 5 = 9$ already, and $3 + 6 = 9$ likewise, leaving no room for a third positive gap).

Now decide the order. The two-step sums $a + b$ and $b + c$ must be the remaining differences $4$ and $6$. If $5$ sat in the middle, one two-step sum would be $5 + 1 = 6$ and the other $5 + 3 = 8$, which is not allowed. So the middle gap $b = 1$, and $(a, c) = (3, 5)$ or $(5, 3)$; both give two-step sums $4$ and $6$.

• Gaps $3, 1, 5$: the numbers are $z + 9, z + 6, z + 5, z$, sum $4z + 20 = 44$, so $z = 6$ and $w = 15$.
• Gaps $5, 1, 3$: the numbers are $z + 9, z + 4, z + 3, z$, sum $4z + 16 = 44$, so $z = 7$ and $w = 16$.

Both lists ($15, 12, 11, 6$ and $16, 11, 10, 7$) really do have differences $1, 3, 4, 5, 6, 9$. The sum of the possible values of $w$ is $15 + 16 = 31$.

The answer is $\boxed{\textbf{(B)}\ 31}$.