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2003 AMC 10A 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) \ } 2003\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+\ldots+4006)-(1+3+5+\ldots+4005).

\displaystyle (2+4+6+\ldots+4006)-(1+3+5+\ldots+4005) = (2-1)+(4-3)+(6-5)+\ldots+(4006-4005)
= 1+1+1+\ldots+1 = 2003 \Longrightarrow \mbox{D}.

Alternatively, using the sum of an arithmetic progression formula, we can write \frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = 2003.

See also

2003 AMC 10A (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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