Difference between revisions of "2016 AMC 10B Problems/Problem 3"

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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>
 
<math>\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048</math>
  
==Solution==
+
==Solution 1==
 
Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math>
 
Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math>
  

Revision as of 02:31, 16 February 2021

Problem

Let $x=-2016$. What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$ ?

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

Solution 1

Substituting carefully, $\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)$

becomes $|4032-2016|+2016=2016+2016=4032$ which is $\boxed{\textbf{(D)}}$.

Solution 2

Solution by e_power_pi_times_i

Substitute $-y = x = -2016$ into the equation. Now, it is $\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y$. Since $y = 2016$, it is a positive number, so $|y| = y$. Now the equation is $\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y$. This further simplifies to $2y-y+y = 2y$, so the answer is $\boxed{\textbf{(D)}\ 4032}$

Video Solution

https://youtu.be/PuAY2v0bt-0

~savannahsolver

See Also

2016 AMC 10B (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 10 Problems and Solutions

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