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2003 AMC 12A Problems/Problem 1: Difference between revisions

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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>
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math>


== Solution ==
==Solution 1==
===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>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math>
<math>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math>


===Solution 2===
The answer is 2003
 
==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>.
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>.






===Solution 3===
==Solution 3==
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).
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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<math>S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow</math> <math>\boxed{\mathrm{(D)}\ 2003}</math>.
<math>S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow</math> <math>\boxed{\mathrm{(D)}\ 2003}</math>.


===Solution 4===
==Solution 4==
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>.
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>.
Solution by franzliszt


==Solution 5 (Fastest method)==
==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 <math>1 * 2003</math> which is <math>\boxed{\mathrm{(D)}\ 2003}</math>.  
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>.


Solution by Penguin Spellcaster
<3


== See also ==
== See also ==
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[[Category:Introductory Algebra Problems]]
[[Category:Introductory Algebra Problems]]
{{MAA Notice}}
{{MAA Notice}}
https://www.youtube.com/watch?v=6ZRnm_DGFfY
Video solution by canada math

Latest revision as of 20:48, 3 November 2025

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

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

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

The answer is 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

These problems are copyrighted © by the Mathematical Association of America.

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

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