Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2016 AMC 12B Problems/Problem 3: Difference between revisions

← Older edit
Mathmaster2012 (talk | contribs)
editor
185 edits
No edit summary
Hwdaniel (talk | contribs)
editor
16 edits
 
(6 intermediate revisions by 4 users not shown)
Line 3: Line 3:
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==
<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math>
 
==Solution 1==
By: dragonfly
 
First of all, lets plug in all of the <math>x</math>'s into the equation.
 
<math>\bigg|</math> <math>||-2016|-(-2016)|-|-2016|</math> <math>\bigg|</math> <math>-(-2016)</math>
 
Then we simplify to get
 
<math>\bigg|</math> <math>|2016+2016|-2016</math> <math>\bigg|</math> <math>+2016</math>
 
which simplifies into
 
<math>\bigg|</math> <math>2016</math> <math>\bigg|</math> <math>+2016</math>
 
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>
 
<math>=4032 \implies \boxed{\textbf{(D)}}</math>
 
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 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 (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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_3&oldid=135059"