Difference between revisions of "2003 AMC 12A Problems/Problem 1"
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===Solution 2=== | ===Solution 2=== | ||
Using the sum of an [[arithmetic progression]] formula, we can write this as <math>\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}</math>. | Using the sum of an [[arithmetic progression]] formula, we can write this as <math>\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}</math>. | ||
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+ | ===Solution 3=== | ||
+ | The formula for the sum of the first <math>n</math> even numbers, is <math>S_E=n^{2}+n</math>, (E standing for even). | ||
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+ | Sum of first <math>n</math> odd numbers, is <math>S_O=n^{2}</math>, (O standing for odd). | ||
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+ | Knowing this, plug <math>2003</math> for <math>n</math>, | ||
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+ | <math>S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow</math> <math>\boxed{\mathrm{(D)}\ 2003}</math>. | ||
== See also == | == See also == |
Revision as of 17:04, 28 December 2011
- The following problem is from both the 2003 AMC 12A #1 and 2003 AMC 10A #1, so both problems redirect to this page.
Problem
What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?
Solution
Solution 1
The first even counting numbers are .
The first odd counting numbers are .
Thus, the problem is asking for the value of .
Solution 2
Using the sum of an arithmetic progression formula, we can write this as .
Solution 3
The formula for the sum of the first even numbers, is , (E standing for even).
Sum of first odd numbers, is , (O standing for odd).
Knowing this, plug for ,
.
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 |
2003 AMC 12A (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 12 Problems and Solutions |