Difference between revisions of "2003 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | The first <math>2003</math> even counting numbers are <math>2,4,6,...,4006</math>. | + | *The first <math>2003</math> even counting numbers are <math>2,4,6,...,4006</math>. |
| + | *The first <math>2003</math> odd counting numbers are <math>1,3,5,...,4005</math>. | ||
| − | + | Thus, the problem is asking for the value of <math>(2+4+6+\ldots+4006)-(1+3+5+\ldots+4005)</math>. | |
| + | :<math>\displaystyle (2+4+6+\ldots+4006)-(1+3+5+\ldots+4005) = (2-1)+(4-3)+(6-5)+\ldots+(4006-4005)</math> | ||
| + | :<math>= 1+1+1+\ldots+1 = 2003 \Longrightarrow \mbox{D}</math>. | ||
| − | + | 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>. | |
| − | + | == See also == | |
| − | + | {{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}} | |
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| − | == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 16:30, 26 February 2007
Problem
What is the difference between the sum of the first 
even counting numbers and the sum of the first odd counting numbers?
Solution
- The first
even counting numbers are 
. - The first odd counting numbers are
.
.
.Thus, the problem is asking for the value of 
.
.
.Alternatively, using the sum of an arithmetic progression formula, we can write 
.
See also
| 2003 AMC 10A (Problems • Answer Key • 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 | ||
| All AMC 10 Problems and Solutions | ||
