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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+...+4006)-(1+3+5+...+4005)</math>.
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| − | <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+\ldots+1 = 2003 \Rightarrow \text{D}</math>
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| − | == See Also ==
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| − | *[[2003 AMC 10A Problems]]
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| − | *[[2003 AMC 10A Problems/Problem 2|Next Problem]]
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| − | [[Category:Introductory Algebra Problems]]
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