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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?  
  
 
<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>
  
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== 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>  
  
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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 by franzliszt
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==Solution 5 (Fastest method)==
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We can see that the difference of the first even number and the first odd number is one, the difference between the second even number and the second odd number is one and so on. Then, we get <math>1 * 2003</math> which is <math>\boxed{\mathrm{(D)}\ 2003}</math>.
  
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== See Also ==
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Solution by Penguin Spellcaster
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*[[2003 AMC 12A Problems]]
 
  
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{{AMC12 box|year=2003|before=First<br />Question|num-a=2}}
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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

Revision as of 20:03, 1 May 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 by franzliszt

Solution 5 (Fastest method)

We can see that the difference of the first even number and the first odd number is one, the difference between the second even number and the second odd number is one and so on. Then, we get $1 * 2003$ which is $\boxed{\mathrm{(D)}\ 2003}$.

Solution by Penguin Spellcaster

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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