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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(n-1)/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(n-1)/2 x = 2000/n - (n-1)/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, n-1 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 - (n-1)/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
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Re: Tough Algebra question

Quote:
 Originally Posted by tiba I'm not 100% sure on the second part, but I made as follow: nx = 2000 - n(n-1)/2 x = 2000/n - (n-1)/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, n-1 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 - (n-1)/2 which give us an integer, but I let this for more advanced mathematicians. Well, that was my try
1+2+...+2n+1 is always divisible by 2n+1, since it equals ((2n+1)(2n+2)/2, which equals (n+1)*(2n+1). But note that ANY sequence of 2n+1 consecutive integers will have the same set of residues mod 2n+1 as the first such consecutive integers, just in various orders. So a number k can be the sum of 2n+1 consecutive integer only if 2n+1 is a factor of k. But as you note in the case of 125, you have to be a factor but also "small enough".

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
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Re: Tough Algebra question

Quote:
 Originally Posted by tiba 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(n-1)/2.
Could you elaborate how you found the formula $\frac{n(n-1)}{2}$?

 June 29th, 2012, 12:06 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 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(=n-k) + [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
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Re: Tough Algebra question

Quote:
 Originally Posted by jorgerocha Could you elaborate...
I believe tiba meant to use:

$x+(x+1)+(x+2)+\cdots+$$x+(n-1)$$=2000$

$nx+\frac{n(n-1)}{2}=2000$

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:

$S=1+2+3+\cdots+(n-3)+(n-2)+(n-1)$

$S=(n-1)+(n-2)+(n-3)+\cdots+3+2+1$

Adding the two equations term by term, we find:

$2S=n+n+n+\cdots+n+n+n$

On the right we have (n-1) terms, thus:

$2S=n(n-1)$

$S=\frac{n(n-1)}{2}$

 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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