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

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becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>.
 
becomes <math>|4032-2016|+2016=2016+2016=4032</math> which is <math>\textbf{(D)}</math>.
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==Solution 2==
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Solution by e_power_pi_times_i
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Substitute <math>-y = x = -2016</math> into the equation. Now, it is <math>\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y</math>. Since <math>y = 2016</math>, it is a positive number, so <math>|y| = y</math>. Now the equation is <math>\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y</math>. This further simplifies to <math>2y-y+y = 2y</math>, so the answer is <math>\textbf{(D)} 4032</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}}
 
{{AMC10 box|year=2016|ab=B|num-b=2|num-a=4}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:12, 21 June 2016

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

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

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

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 $\textbf{(D)} 4032$

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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