2016 AMC 10B Problem 3

P1-10~1 minPrint

Absolute Value & Inequalities Set up the equation/formula and compute; no special trick needed

Let $x=-2016$ . What is the value of $\bigg| \big||x|-x\big|-|x| \bigg| -x?$

Show solution without answering

Key insight

Since x is negative, both |x| and -x equal 2016. Replace them with 2016 and work from the innermost bars outward.

Solution

The whole difficulty is signs, so first write down two facts about $x = -2016$ .

$|x| = 2016$ , because absolute value removes the minus sign.
$-x = -(-2016) = +2016$ . So "minus $x$ " really means "plus $2016$ ."

Now work from the inside of the bars to the outside.

Innermost: $|x| - x = 2016 + 2016 = 4032$ . Its absolute value is still $4032$ .
Next: $4032 - |x| = 4032 - 2016 = 2016$ . Its absolute value is still $2016$ .
Last: $2016 - x = 2016 + 2016 = 4032$ .

The answer is $\boxed{\textbf{(D)}\ 4032}$ .

Why this works

Nested absolute values are handled like nested parentheses: start in the middle and move out. Here every number inside the bars turns out positive, so the bars change nothing. The only real danger is the sign of $-x$ , so settle that before you start.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since $x = -2016$ is negative, $|x| = 2016$ and $-x = 2016$ . Work from the innermost bars outward.

$|x| - x = 2016 + 2016 = 4032$ .
$\big| 4032 \big| = 4032$ .
$4032 - |x| = 4032 - 2016 = 2016$ .
$\big| 2016 \big| = 2016$ .
Finally, $2016 - x = 2016 + 2016 = 4032$ .

The answer is $\boxed{\textbf{(D)}\ 4032}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)
Intermediate Algebra, Chapter 12: Inequalities (§12.3, §12.4)