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

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Let <math>x=-2016</math>. What is the value of <math>\bigg|</math> <math>||x|-x|-|x|</math> <math>\bigg|</math> <math>-x</math>?
 
Let <math>x=-2016</math>. What is the value of <math>\bigg|</math> <math>||x|-x|-|x|</math> <math>\bigg|</math> <math>-x</math>?
  
==Solution==
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<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math>
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==Solution 1==
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By: dragonfly
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 +
First of all, lets plug in all of the <math>x</math>'s into the equation.
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<math>\bigg|</math> <math>||-2016|-(-2016)|-|-2016|</math> <math>\bigg|</math> <math>-(-2016)</math>
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Then we simplify to get
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<math>\bigg|</math> <math>|2016+2016|-2016</math> <math>\bigg|</math> <math>+2016</math>
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which simplifies into
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<math>\bigg|</math> <math>2016</math> <math>\bigg|</math> <math>+2016</math>
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and finally we get <math>\boxed{\textbf{(D)}\ 4032}</math>
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==Solution 2==
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Consider <math>x</math> is negative.
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We replace all instances of <math>x</math> with <math>|x|</math>:
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<math>\bigg|</math> <math>||x|+|x||-|x|</math> <math>\bigg|</math> <math>+|x|</math>
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<math>=</math> <math>\bigg|</math> <math>|2x|-|x|</math> <math>\bigg|</math> <math>+|x|</math>
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<math>=</math> <math>|x|</math> <math>+|x|</math>
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<math>=|2x|</math>
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<math>=4032 \implies \boxed{\textbf{(D)}}</math>
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omerselim1
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}}
 
{{AMC12 box|year=2016|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:20, 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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