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Sum of n consecutive numbers is divisible by n

  1. Jan 18, 2015 #1
    I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n.

    I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.
     
  2. jcsd
  3. Jan 18, 2015 #2

    lavinia

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    try writing down a general formula for the sum
     
  4. Jan 18, 2015 #3

    mathman

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    The remainders, after dividing by n, for n consecutive numbers are 0,1,...,n-1. Sum the remainders to get n(n-1)/2.

    If n is odd, n-1 is even and (n-1)/2 is an integer, so this sum is divisible by n.

    If n is even, n-1 is odd. n(n-1)/2 = n(n-2)/2 + n/2. (n-2)/2 is an integer, so n/2 is the remainder after dividing the sum by n.
     
    Last edited: Jan 19, 2015
  5. Jan 22, 2015 #4

    HallsofIvy

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    Another way to look at it: to find the sum of numbers from k to k+ n- 1, find the sum from 1 to k+ n- 1, (k+ n- 1)(k+ n)/2, then subtract the sum from 1 to k, k(k- 1)/2. That is ((k^2+ 2kn+ n^2- k- n)- (k^2- k))/2= (2kn+ n^2- n)/2= n((2k- n- 1)/2) so is divisible by n.
     
  6. Jan 22, 2015 #5

    mathman

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    You missed the point. In your derivation 2k-n-1 must be even, so n must be odd. When n is even the expression has a remainder of n/2.
     
  7. Feb 9, 2015 #6
    Thank you very much for your everyone. I really appreciate it.

    Mathman, could I run a couple of problems past you if you have a moment?

    Regards,
    Matt.
     
  8. Feb 9, 2015 #7
    Thank you very much for your everyone. I really appreciate it.
    Mathman, could I run a couple of problems past you if you have a moment?

    Regards,
    Matt.
     
  9. Feb 9, 2015 #8

    mathman

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    If you submit them via the forum I will look at them. I prefer not to answer privately.
     
  10. Feb 10, 2015 #9

    HallsofIvy

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    How do you get that?

     
  11. Feb 10, 2015 #10

    mathman

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    In your derivation you assumed n((2k- n- 1)/2) is an integer divisible by n. Therefore (2k-1-n)/2 must be an integer.

    2k--n-1: 2k is even, if n is even, n+1 is odd. Even number minus odd number is odd.
     
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