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 16:49, 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.
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 | |
