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June 29th, 2012, 01:03 AM  #1 
Newbie Joined: Jun 2012 Posts: 7 Thanks: 0  Tough Algebra question
Hi, I'm new to this forum and I'm stuck on a maths problem. Any ideas would be appreciated: The sum of a certain number of consecutive positive integers is 2000. Find an example of these numbers. Is this solution unique? If not, how many solutions are there? 
June 29th, 2012, 01:35 AM  #2 
Member Joined: Jan 2012 Posts: 52 Thanks: 0  Re: Tough Algebra question
The answer is given by: x + (x + 1) + (x + 2) + (x + 3) ... (x + n) = 2000 Which turns out to be: n.x + S = 2000 Where n is the number of numbers you want to use and S is given by the formula n(n1)/2. For example, for n=5, we have 5x + 10 = 2000 x = 398 So one of the possibilities is 398, 399, 400, 401 and 402, which sum up to 2000. This is the first part of the problem 
June 29th, 2012, 02:08 AM  #3 
Member Joined: Jan 2012 Posts: 52 Thanks: 0  Re: Tough Algebra question I'm not 100% sure on the second part, but I made as follow: nx = 2000  n(n1)/2 x = 2000/n  (n1)/2 Since x is an integer, n must be from the divisors of 2000, given below: 1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 1000, 2000. Also, n1 must be an even number, that is, n must be an odd number. The only two that satisfies our conditions are 5 and 25 (since 125.124/2 is greater than 2000). For 25, we have these numbers, which sum up to 2000: 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91 and 92. My only doubt is if there isn't two rational numbers 2000/n  (n1)/2 which give us an integer, but I let this for more advanced mathematicians. Well, that was my try 
June 29th, 2012, 03:13 AM  #4  
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Tough Algebra question Quote:
Odd composite numbers are very well behaved re sums of consecutive integers. Any composite odd number 2k+1 that is the product of odd numbers 2x+1 and 2y+1, where 2x+1 ? 2y+1 is the sum of 2y+1 integers that are "centered" on 2x+1 and vice versa, eg 21 = 6+7+8. Actually, it works straightforwardly the other way around, too, in many cases, eg 35 = 5+6+7+8+9 = 2+3+4+5+6+7+8. It will ALWAYS work in both directions if you allow yourself to start at 0 or even negative numbers, eg 33= 10+11+12=(2)+(1)+0+1+2+3+4+5+6+7+8  
June 29th, 2012, 09:51 AM  #5  
Newbie Joined: Jun 2012 Posts: 5 Thanks: 0  Re: Tough Algebra question Quote:
 
June 29th, 2012, 12:06 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,590 Thanks: 1038  Re: Tough Algebra question
Also valid (32 terms) : 47+48+49+...+77+ 78 = 2000 And of course the 1term 2000 Using a simple example: number of terms added = k = 3 ; sum = S = 24 1 + 2 + 3 + 4 + 5 + 6(=nk) + [7 + 8 + 9(=n)] n(n +1) / 2  (n  k)(n  k + 1)/2 = S leads to: k = [2n + 1 + SQRT(4n^2 + 4n + 1  8S)] / 2 That'll give k=16 (+ solution) and k=3 ( solution) as solutions. The "16" solution, I think, represents: 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 ; 6 to 6 = 0, 7+8+9 = 24 
June 29th, 2012, 12:13 PM  #7  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Tough Algebra question Quote:
There are many ways to derive the formula for the summation of the first n  1 natural numbers, but perhaps the simplest is to write: Adding the two equations term by term, we find: On the right we have (n1) terms, thus:  
June 29th, 2012, 02:12 PM  #8 
Member Joined: Jan 2012 Posts: 52 Thanks: 0  Re: Tough Algebra question
So, what is the final answer, I'm curious 
June 29th, 2012, 02:38 PM  #9 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Tough Algebra question
The sum of 2n consecutive integers will always be n mod 2n.

June 29th, 2012, 03:18 PM  #10 
Newbie Joined: Jun 2012 Posts: 7 Thanks: 0  Re: Tough Algebra question
Damn it, I was convinced that n=5 and n=25 were the only two possible solutions, ignoring n=1. Now there is n=32. So does anyone have an idea how many possible solutions there are and why?


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