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Difference between revisions of "2003 AMC 10A Problems/Problem 1"
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<math>(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) </math>  
 
<math>(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) </math>  
  
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<math>= 1+1+1+...+1 = 2003 \Rightarrow D</math>
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<math>= 1+1+1+\ldots+1 = 2003 \Rightarrow \text{D}</math>
  
 
== See Also ==
 
== See Also ==

Revision as of 18:42, 22 November 2006

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

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

$= 1+1+1+\ldots+1 = 2003 \Rightarrow \text{D}$

See Also

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