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Difference between revisions of "2010 AMC 12A Problems/Problem 1"
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== Solution ==
 
== Solution ==
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<math>20-2010+201+2010-201+20</math>
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<math>20-2010+201+2010-201+20=20+20=40; \boxed{\textbf{(C)}}</math>.
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<math>20+20</math>
 
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<math>\boxed{\textbf{(C)}\ 40}</math>
 

Revision as of 20:49, 12 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

$20-2010+201+2010-201+20=20+20=40; \boxed{\textbf{(C)}}$.

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