2003 AMC 12A Problems/Problem 1

The following problem is from both the 2003 AMC 12A #1 and 2003 AMC 10A #1, so both problems redirect to this page.

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?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$

Solution

Solution 1

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

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

$= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$

Solution 2

Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}$ .

2003 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Question	Followed by Problem 2
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All AMC 12 Problems and Solutions

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2003 AMC 12A Problems/Problem 1

Contents

Problem

Solution

Solution 1

Solution 2

See also