Difference between revisions of "2010 AMC 12A Problems/Problem 1"

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<math>\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020</math>
 
<math>\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020</math>
  
[[2010 AMC 12A Problems/Problem 1|Solution]]
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== Solution ==
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<math>20-2010+201+2010-201+20</math>
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<math>20+20</math>
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<math>\boxed{\textbf{(C)}\ 40}</math>

Revision as of 15:37, 10 February 2010

Problem 1

What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$?

$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$

Solution

$20-2010+201+2010-201+20$

$20+20$

$\boxed{\textbf{(C)}\ 40}$