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| − | == Problem ==
| + | #REDIRECT[[2003 AMC 12A Problems/Problem 1]] |
| − | What is the difference between the sum of the first <math>2003</math> even counting numbers and the sum of the first <math>2003</math> odd counting numbers?
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| − | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math>
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| − | == Solution ==
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| − | *The first <math>2003</math> even counting numbers are <math>2,4,6,...,4006</math>.
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| − | *The first <math>2003</math> odd counting numbers are <math>1,3,5,...,4005</math>.
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| − | Thus, the problem is asking for the value of <math>(2+4+6+\ldots+4006)-(1+3+5+\ldots+4005)</math>.
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| − | :<math>\displaystyle (2+4+6+\ldots+4006)-(1+3+5+\ldots+4005) = (2-1)+(4-3)+(6-5)+\ldots+(4006-4005)</math>
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| − | :<math>= 1+1+1+\ldots+1 = 2003 \Longrightarrow \mbox{D}</math>.
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| − | Alternatively, using the sum of an [[arithmetic progression]] formula, we can write <math>\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = 2003</math>.
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| − | == See also ==
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| − | {{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}}
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| − | [[Category:Introductory Algebra Problems]]
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