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

Revision as of 21:10, 25 January 2018

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

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

2016 AMC 10B (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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 Substituting carefully, <math>\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)</math>
-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>\boxed{\textbf{(D)}}</math>.
 ==Solution 2==

Page

Toolbox

Search

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

Revision as of 21:10, 25 January 2018

Contents

Problem

Solution

Solution 2

See Also