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2003 AMC 12A Problems/Problem 1

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Problem

What is the Difference between the sum of the first 2003 even counting numbers and the sum of the first 2003 odd counting numbers?

\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2006\qquad \mathrm{(E) \ } 4006

Solution

The first 2003 even counting numbers are 2,4,6,...,4006.

The first 2003 odd counting numbers are 1,3,5,...,4005.

Thus, the problem is asking for the value of (2+4+6+...+4006)-(1+3+5+...+4005).

(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)

= 1+1+1+...+1 = 2003 \Rightarrow D

See Also

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Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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