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A solution to this problem?

  1. Jul 21, 2004 #1
    Replace some elements of the natural sequence 1, 2, 3, ... such that after this replacement the sum of first k terms is divisible by k.

    Any solutions?
     
  2. jcsd
  3. Jul 21, 2004 #2

    loseyourname

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    Replace every element from 2 onward with 1. Then the sum of the first k terms will be k.
     
  4. Jul 22, 2004 #3
    Replace all elements (except the k:th) with 0. Let the k:th element be k (or ak for some integer a ;)).
     
  5. Jul 22, 2004 #4

    NateTG

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    What if by some he means 'a finite subset'?
     
  6. Jul 22, 2004 #5

    Njorl

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    Doubling every number works.

    sum(1 to k)=k(k+1)/2 so multiplying this by 2 yields k(k+1) which is of course divisible by k.

    Still, that is replacing all of the numbers, but I think it is on the road to a more pleasing solution.

    Njorl
     
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