Difference between revisions of "2003 AMC 12A Problems/Problem 1"
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+ | {{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #1]] and [[2003 AMC 10A Problems|2003 AMC 10A #1]]}} | ||
== Problem == | == Problem == | ||
− | What is the | + | 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? |
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math> | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math> | ||
== Solution == | == Solution == | ||
+ | ===Solution 1=== | ||
+ | |||
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>. | ||
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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> | ||
− | <math>= 1+1+1+...+1 = | + | <math>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math> |
− | == | + | ===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>. | |
+ | == See also == | ||
+ | {{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}} | ||
{{AMC12 box|year=2003|ab=A|before=First Question|num-a=2}} | {{AMC12 box|year=2003|ab=A|before=First Question|num-a=2}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 15:49, 29 July 2011
- The following problem is from both the 2003 AMC 12A #1 and 2003 AMC 10A #1, so both problems redirect to this page.
Contents
[hide]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 .
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 |