2011 AMC 10B Problem 21

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 $\boxed{\textbf{(B)}\ 31}$.

For an ordered list, every pairwise difference is a sum of consecutive gaps, and the largest difference is the total spread. Pinning down the extreme difference first collapses the search to choosing three gaps with a fixed sum, and the two-step sums then test each ordering. The final sum condition is a one-variable linear equation.

Since $w - z = 9$ and $x, y$ lie strictly between, write $w = z + 9$ and note $x + y = 44 - 2z - 9 = 35 - 2z$. Trying small $z$ and listing which pairs $(x, y)$ inside $(z, z + 9)$ produce exactly the six required differences quickly finds $z = 6$ and $z = 7$ and nothing else.

• Stopping after finding one valid arrangement of the gaps ($w = 16$ or $w = 15$) and reporting that single value instead of the sum of both.
• Forgetting that the middle gap must combine with both neighbors to give listed differences, and accepting an order like $1, 3, 5$ whose two-step sum $8$ is not in the list.
• Assuming the differences $1, 3, 4, 5, 6, 9$ are listed in some order that corresponds to positions, rather than treating them as an unordered set.