Difference between revisions of "2016 AMC 12B Problems/Problem 3"

Revision as of 10:43, 5 November 2019

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

$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

By: dragonfly

First of all, lets plug in all of the $x$ 's into the equation.

$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$

Then we simplify to get

$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$

which simplifies into

$\bigg|$ $2016$ $\bigg|$ $+2016$

and finally we get $\boxed{\textbf{(D)}\ 4032}$

Consider $x$ is negative.

We replace all instances of $x$ with $|x|$ :

$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$

$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$

$=$ $|x|$ $+|x|$

$=|2x|$ =4032 \boxed{\textbf{(D)}\}$

omerselim1

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 5: / Line 5: @@
 <math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math>
-==Solution==
+==Solution 1==
 By: dragonfly
@@ Line 21: / Line 21: @@
 and finally we get <math>\boxed{\textbf{(D)}\ 4032}</math>
+==Solution 2==
+Consider <math>x</math> is negative.
+We replace all instances of <math>x</math> with <math>|x|</math>:
+<math>\bigg|</math> <math>||x|+|x||-|x|</math> <math>\bigg|</math> <math>+|x|</math>
+<math>=</math> <math>\bigg|</math> <math>|2x|-|x|</math> <math>\bigg|</math> <math>+|x|</math>
+<math>=</math> <math>|x|</math> <math>+|x|</math>
+<math>=|2x|
+</math>=4032 \boxed{\textbf{(D)}\}$
+omerselim1
 ==See Also==
 {{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}