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

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<math>=|2x|</math>
 
<math>=|2x|</math>
  
<math>=4032 \boxed{\textbf{(D)}}</math>
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<math>=4032 \implies \boxed{\textbf{(D)}}</math>
  
 
omerselim1
 
omerselim1

Latest revision as of 21:20, 14 October 2020

Problem

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

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

Solution 1

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 $\boxed{\textbf{(D)}\ 4032}$

Solution 2

Consider $x$ is negative.

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

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

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

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

$=|2x|$

$=4032 \implies \boxed{\textbf{(D)}}$

omerselim1

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
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

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