2009 AMC 10A Problem 16

Let

a,

b,

c,

and

d

be real numbers with

|a - b| = 2,

|b - c| = 3,

and

|c - d| = 4.

What is the sum of all possible values of

|a - d|?

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Solution

Since

a - d

= (a-b) + (b-c) + (c-d)

= \pm 2 \pm 3 \pm 4,

the possible absolute values are

\begin{aligned} 2+3+4 &= 9, \\ 2+3-4 &= 1, \\ 2-3+4 &= 3, \\ -2+3+4 &= 5. \end{aligned}

Their sum is

9 + 1 + 3 + 5 = 18.

Thus, the correct answer is D.

Tagged: absolute value · casework