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

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==Solution==
 
==Solution==
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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>
  
 
==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}}

Revision as of 11:08, 21 February 2016

Problem

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

Solution

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

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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