# number of series possible

• Oct 31st 2013, 12:40 AM
nikhil
number of series possible
Greetings,
struck with the question that how many sequences of consecutive numbers exist such that on adding them we get 500.

thanks
• Oct 31st 2013, 04:03 AM
Plato
Re: number of series possible
Quote:

Originally Posted by nikhil
The question that how many sequences of consecutive numbers exist such that on adding them we get 500.

You need to find how many integer pairs $\displaystyle (n,x)$ satisfy $\displaystyle \sum\limits_{k = 0}^n {(x + k)}=500~.$

That can be written as $\displaystyle \left( {n + 1} \right)x + \frac{{n(n + 1)}}{2} = 500$.
• Oct 31st 2013, 06:04 AM
nikhil
Re: number of series possible
Thanks Plato but am not able to find the integral solutions to this equation as x is itself a variable(1st term of sequence)
• Oct 31st 2013, 06:54 AM
SlipEternal
Re: number of series possible
Assume $\displaystyle x$ is a constant and solve for $\displaystyle n$:

$\displaystyle n = \dfrac{1}{2}\left(\sqrt{4x^2-4x+4001}-2x-1\right)$

Since both $\displaystyle x$ and $\displaystyle n$ must be integers, figure out for which values of $\displaystyle x$ $\displaystyle 4x^2-4x+4001$ is an odd perfect square (it needs to be odd since otherwise, $\displaystyle n$ will not be an integer).

Edit: Also $\displaystyle n$ must be nonnegative and $\displaystyle x\le 500$... That leaves only 8 solutions.

Here are the first two:
$\displaystyle (n,x) = (0,500)$ or $\displaystyle (n,x) = (999,-499)$.

Every solution $\displaystyle (n,x)$ with $\displaystyle x>0$ will have another solution $\displaystyle (n+2x-1,1-x)$ since adding all of the numbers from $\displaystyle 1-x$ to $\displaystyle x-1$ will give zero, then you just have the same $\displaystyle n$ integers from $\displaystyle x$ to $\displaystyle x+n-1$ from the initial solution. Hence, if you find three more solutions with $\displaystyle x>0$, then you have found all of the solutions.
• Oct 31st 2013, 08:39 AM
ebaines
Re: number of series possible
Yes, 8 solutions. To be honest I would have said only 4 solutions, not realizing until reading SlipEternal's post that each solution a, a+1, a+2,...a+n has a counterpart of -(a-1), -(a-2), -(a-3) ... a-3, a-2, a-1, a, a+1, a+2,...a+n. My approach is to check the value of the median number in the series of n integers to see if it "makes sense" - for n odd the median number must be an integer, and for n even the mediian must be an integer + 1/2. This is because the starting number of the sequence of n terms is (500/n)-(n-1)/2. If this is an integer you have a sequence of n terms that adds to 500. I found four values for n that work, so using SlipEternal's idea that means there are a total of 8.
• Oct 31st 2013, 11:26 AM
nikhil
Re: number of series possible
thanks slipEternal and ebaines. SlipEternal could you plz explain how you got the equation will be odd perfect square 8 times or just give any online reference ( lyk Diophantine equation or smthin)
• Oct 31st 2013, 11:43 AM
SlipEternal
Re: number of series possible
ebaines idea is easier to use. Solve for $\displaystyle x$ instead of solving for $\displaystyle n$. Then you have $\displaystyle x = \dfrac{500}{n+1} - \dfrac{n}{2}$. Now, let's consider cases where $\displaystyle x$ is an integer.

Case 1: 2 divides n
Then (n+1) must divide 500. Since (n+1) must be odd, we know $\displaystyle n+1 \in \{1,5,25,125\}$ as those are the only odd factors of 500.

Case 2: 2 does not divide n.
Then n must be odd and n+1 must be even. But, n+1 cannot divide 500. It must divide 1,000, though. Moreover, $\displaystyle \dfrac{1000}{n+1}$ must be odd. Since the only odd factors of 1,000 are the odd factors of 500, we have $\displaystyle \dfrac{1000}{n+1} \in \{1,5,25,125\}$.

Those are all eight solutions.