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 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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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 ==
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===Solution 1===
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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 = 2003 \Rightarrow D</math>
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<math>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math>
  
== See Also ==
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===Solution 2===
*[[2003 AMC 12A Problems]]
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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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== See also ==
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{{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.

Problem

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$

Solution

Solution 1

The first $2003$ even counting numbers are $2,4,6,...,4006$.

The first $2003$ odd counting numbers are $1,3,5,...,4005$.

Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$.

$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$

$= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$

Solution 2

Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}$.

See also

2003 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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
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