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Math Help - number of series possible

  1. #1
    Senior Member nikhil's Avatar
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    Question 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
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  2. #2
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    Re: number of series possible

    Quote Originally Posted by nikhil View Post
    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 (n,x) satisfy \sum\limits_{k = 0}^n {(x + k)}=500~.

    That can be written as \left( {n + 1} \right)x + \frac{{n(n + 1)}}{2} = 500.
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  3. #3
    Senior Member nikhil's Avatar
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    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)
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    Re: number of series possible

    Assume x is a constant and solve for n:

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

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

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

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

    Every solution (n,x) with x>0 will have another solution (n+2x-1,1-x) since adding all of the numbers from 1-x to x-1 will give zero, then you just have the same n integers from x to x+n-1 from the initial solution. Hence, if you find three more solutions with x>0, then you have found all of the solutions.
    Last edited by SlipEternal; October 31st 2013 at 07:18 AM.
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  5. #5
    MHF Contributor ebaines's Avatar
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    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.
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  6. #6
    Senior Member nikhil's Avatar
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    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)
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    Re: number of series possible

    ebaines idea is easier to use. Solve for x instead of solving for n. Then you have x = \dfrac{500}{n+1} - \dfrac{n}{2}. Now, let's consider cases where x is an integer.

    Case 1: 2 divides n
    Then (n+1) must divide 500. Since (n+1) must be odd, we know 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, \dfrac{1000}{n+1} must be odd. Since the only odd factors of 1,000 are the odd factors of 500, we have \dfrac{1000}{n+1} \in \{1,5,25,125\}.

    Those are all eight solutions.
    Last edited by SlipEternal; October 31st 2013 at 11:45 AM.
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