2010 AMC 12A Problems/Problem 1

Revision as of 19:49, 12 February 2010 by AIME15 (talk | contribs) (Solution)

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

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