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 | + | 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) \ } | + | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 </math> |
+ | |||
+ | ==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>(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> | + | <math>= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}</math> |
+ | |||
+ | ==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>. | ||
+ | |||
+ | |||
+ | |||
+ | ==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). | ||
+ | |||
+ | Sum of first <math>n</math> odd numbers, is <math>S_O=n^{2}</math>, (O standing for odd). | ||
+ | |||
+ | Knowing this, plug <math>2003</math> for <math>n</math>, | ||
+ | |||
+ | <math>S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow</math> <math>\boxed{\mathrm{(D)}\ 2003}</math>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ==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 <math>1 \cdot 2003</math>, and the answer is <math>\boxed{\text{(D) }2003}</math>. | ||
− | + | <3 | |
− | |||
− | + | == See also == | |
+ | {{AMC10 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]] | ||
+ | {{MAA Notice}} | ||
+ | https://www.youtube.com/watch?v=6ZRnm_DGFfY | ||
+ | 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.
Contents
[hide]Problem
What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?
Solution 1
The first even counting numbers are .
The first odd counting numbers are .
Thus, the problem is asking for the value of .
Solution 2
Using the sum of an arithmetic progression formula, we can write this as .
Solution 3
The formula for the sum of the first even numbers, is , (E standing for even).
Sum of first odd numbers, is , (O standing for odd).
Knowing this, plug for ,
.
Solution 4
In the case that we don't know if is considered an even number, we note that it doesn't matter! The sum of odd numbers is . And the sum of even numbers is either or . When compared to the sum of odd numbers, we see that each of the th term in the series of even numbers differ by . For example, take series and . The first terms are and . Their difference is . Similarly, take take series and . The first terms are and . Their difference is . Since there are terms in each set, the answer .
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 , and the answer is .
<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.
https://www.youtube.com/watch?v=6ZRnm_DGFfY Video solution by canada math