Difference between revisions of "2003 AMC 10A Problems/Problem 1"

m
(edit solution, box)
Line 5: Line 5:
  
 
== 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>.  
  
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>.
  
Thus, the problem is asking for the value of <math>(2+4+6+...+4006)-(1+3+5+...+4005)</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>.
  
<math>(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) </math>
+
== See also ==
 
+
{{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}}
<math>= 1+1+1+\ldots+1 = 2003 \Rightarrow \mbox{D}</math>
 
 
 
== See Also ==
 
*[[2003 AMC 10A Problems]]
 
 
 
*[[2003 AMC 10A Problems/Problem 2|Next Problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 17:30, 26 February 2007

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

  • 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 $(2+4+6+\ldots+4006)-(1+3+5+\ldots+4005)$.

$\displaystyle (2+4+6+\ldots+4006)-(1+3+5+\ldots+4005) = (2-1)+(4-3)+(6-5)+\ldots+(4006-4005)$
$= 1+1+1+\ldots+1 = 2003 \Longrightarrow \mbox{D}$.

Alternatively, using the sum of an arithmetic progression formula, we can write $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = 2003$.

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