Challenging math question (9th grade math)?

what is the diffrence between the sum of the first 2004 positve integers and the sum of the next 2004 postive?


A) 2004 B) 4008 C) 2004^2 D) 4008^2





This was my math homework and i am totally lost but i don't want to loose the points for the problem so can someone please help me. If you can solve this please try to explain how you sound this answer. Thanx

Challenging math question (9th grade math)?
Let's do one a little simpler. Try the difference of the sum of 1,2, %26amp; 3 and 4,5,%26amp; 6. If we take the last number of each set and subtract we get 3. If we take the second numbers: 2 %26amp; 5 and subtract we get 3. And last we take the first numbers 1 %26amp; 4 we get 3. Why 3? Because that is how many numbers are in each set. Your two sets each have 2004 numbers. If we were to take the last set of numbers and place them on top of the first set of numbers and subtract each one we would get 2004 every single time. Since the first set goes to 2004 we know this would happen exactly 2004 times. So we have to add 2004 to 2004... 2004 times. Or more simply put 2004x2004. Or even more simply put 2004^2.
Reply:Hmm...process of elimination...


It's definitely not A. I don't think it's B or D.





(Explanation: It couldn't be a, because for that to be possible every single integer would have to be 1--and that's impossible. It can't be 4008 because if you apply this question to a lower set of numbers (1-10 and 11-20), you'll notice that the sum of all numbers 1 through 10 is 55, which is much greater than the largest number possible--20). Therefore, B can't be the answer...


D can't be the answer because 4008 x 4008 is 16, 064, 064--and that is completely ridiculous.)





I think it's C, but even that seems kinda skeptical.
Reply:A.)





The first 2004 positive integers would be 1002 because there are 50/50 even to odd . Therefore, you would divide 2004 by two.





Then you do the same for the next 2004. Which is 1002





When you add 1002 with 1002, you get 2004.





This seems like the most appropriate answer. I kinda took a guess at the process, so don't count on this answer as correct. But all the other numbers are ridiculously large.





I'm pretty sure that A is the answer.
Reply:Sorry, I misunderstood the question.





idk
Reply:I'd say it's c
Reply:Just imagine the sum of all the numbers to 2004


(as 1,2,3,4....2004)


Then the next numbers as (2005, 2006, 2007, 2008....4008)


Now try to pair off the numbers that are 2004 away from each other (1 with 2005, 2 with 2006...2004 with 4008)


There are going to be 2004 of these pairs, becuase there are 4008 total numbers)


Each of these pairs have a difference of 2004 so the answer is going to be (2004)(2004) or C), 2400^2