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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 ==
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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?  
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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?  
  
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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2006\qquad \mathrm{(E) \ } 4006 </math>
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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math>
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==Solution 1==
  
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== 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>.  
  
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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>  
  
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<math>= 1+1+1+...+1 = 2003 \Rightarrow D</math>
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<math>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math>
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==Solution 2==
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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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==Solution 3==
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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>.
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==Solution 4==
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In the case that we don't know if <math>0</math> is considered an even number, we note that it doesn't matter! The sum of odd numbers is <math>O=1+3+5+...+4005</math>. And the sum of even numbers is either <math>E_1=0+2+4...+4004</math> or <math>E_2=2+4+6+...+4006</math>. When compared to the sum of odd numbers, we see that each of the <math>n</math>th term in the series of even numbers differ by <math>1</math>. For example, take series <math>O</math> and <math>E_1</math>. The first terms are <math>1</math> and <math>0</math>. Their difference is <math>|1-0|=1</math>. Similarly, take take series <math>O</math> and <math>E_2</math>. The first terms are <math>1</math> and <math>2</math>. Their difference is <math>|1-2|=1</math>. Since there are <math>2003</math> terms in each set, the answer <math>\boxed{\mathrm{(D)}\ 2003}</math>.
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==Solution 5 (Fastest method)==
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We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is <math>1 \cdot 2003</math>, and the answer is <math>\boxed{\text{(D) }2003}</math>.
  
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== See Also ==
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<3
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*[[2003 AMC 12A Problems]]
 
  
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*[[2003 AMC 12A/Problem 2|Next Problem]]
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== See also ==
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{{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}}
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{{AMC12 box|year=2003|ab=A|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}
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https://www.youtube.com/watch?v=6ZRnm_DGFfY
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Video solution by canada math

Latest revision as of 19:31, 28 December 2021

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 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 $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}$.


Solution 3

The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$, (E standing for even).

Sum of first $n$ odd numbers, is $S_O=n^{2}$, (O standing for odd).

Knowing this, plug $2003$ for $n$,

$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{\mathrm{(D)}\ 2003}$.

Solution 4

In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$. And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$. When compared to the sum of odd numbers, we see that each of the $n$th term in the series of even numbers differ by $1$. For example, take series $O$ and $E_1$. The first terms are $1$ and $0$. Their difference is $|1-0|=1$. Similarly, take take series $O$ and $E_2$. The first terms are $1$ and $2$. Their difference is $|1-2|=1$. Since there are $2003$ terms in each set, the answer $\boxed{\mathrm{(D)}\ 2003}$.

Solution 5 (Fastest method)

We can pair each term of the sums - the first even number with the first odd number, the second with the second, and so forth. Then, there are 2003 pairs with a difference of 1 in each pair - 2-1 is 1, 4-3 is 1, 6-5 is 1, and so on. Then, the solution is $1 \cdot 2003$, and the answer is $\boxed{\text{(D) }2003}$.

<3

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

https://www.youtube.com/watch?v=6ZRnm_DGFfY Video solution by canada math

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