My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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?
mark777 is offline  
 
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
tiba is offline  
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
tiba is offline  
June 29th, 2012, 03:13 AM   #4
Math Team
 
Joined: Apr 2012

Posts: 1,579
Thanks: 22

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
johnr is offline  
June 29th, 2012, 09:51 AM   #5
Newbie
 
Joined: Jun 2012

Posts: 5
Thanks: 0

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 ?
jorgerocha is offline  
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
Denis is offline  
June 29th, 2012, 12:13 PM   #7
Senior Member
 
MarkFL's Avatar
 
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:
Originally Posted by jorgerocha
Could you elaborate...
I believe tiba meant to use:





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 (n-1) terms, thus:



MarkFL is offline  
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
tiba is offline  
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.
johnr is offline  
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?
mark777 is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
algebra, question, tough



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Modulus question a little tough shikharras Algebra 4 May 11th, 2013 08:44 AM
tough integration question hooperoo Calculus 2 April 24th, 2012 04:23 AM
Why is Math So Boring, And Why is Algebra So Tough? Fire7 Algebra 13 April 24th, 2011 01:00 PM
Tough Probability Question Markthedude Advanced Statistics 2 January 3rd, 2010 07:43 AM
A very tough question cyanide911 Algebra 0 August 5th, 2009 03:26 AM





Copyright © 2019 My Math Forum. All rights reserved.