2009 AMC 10A Problem 16

The three conditions say $a - b = \pm 2$, $b - c = \pm 3$, $c - d = \pm 4$, with the signs chosen independently. Adding,
$$a - d = (a - b) + (b - c) + (c - d) = \pm 2 \pm 3 \pm 4.$$

List the eight sign patterns (the negatives of each pattern give the same absolute value, so four suffice):
- $2 + 3 + 4 = 9$
- $2 + 3 - 4 = 1$
- $2 - 3 + 4 = 3$
- $-2 + 3 + 4 = 5$

The other four patterns are the negatives $-9, -1, -3, -5$. So $|a - d|$ can be $1, 3, 5$ or $9$, and each really occurs (e.g. $a = 0$ and the appropriate $b, c, d$).

Sum: $1 + 3 + 5 + 9 = 18$.

The answer is $\boxed{\textbf{(D)}\ 18}$.

Absolute values hide a sign, and the differences chain together so that $a - d$ is just a signed sum of the three given gaps. Enumerating sign patterns is genuine but tiny casework. Note the symmetry: flipping all signs negates the sum, so only half the patterns need checking.

Think on the number line: start at $a$, step $2$ left or right to $b$, then $3$, then $4$. The distance from the start is $|\pm 2 \pm 3 \pm 4|$; the reachable distances are the odd numbers $1, 3, 5, 9$ (all have the parity of $2 + 3 + 4$), and $7$ is not attainable.

• Assuming a, b, c, d sit in order on the number line so |a-d| = 9, or listing only some sign patterns and missing a value like 1.
• Summing all eight signed values (which cancel to $0$) or all eight absolute values ($36$) instead of the four distinct possible values.