Can Doubling Elements in a Sequence Ensure Divisibility by Their Position?

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The discussion explores whether replacing elements in the natural sequence can ensure that the sum of the first k terms is divisible by k. One proposed solution is to replace every element from 2 onward with 1, resulting in a sum of k. Another suggestion involves replacing all elements except the k-th with 0 and setting the k-th element to k, which also maintains divisibility. Doubling each number in the sequence is mentioned as a valid method, as it results in a sum that is inherently divisible by k. The conversation indicates a search for more elegant solutions while confirming that these approaches work.
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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?
 
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Replace every element from 2 onward with 1. Then the sum of the first k terms will be k.
 
Replace all elements (except the k:th) with 0. Let the k:th element be k (or ak for some integer a ;)).
 
Ben1587 said:
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?

What if by some he means 'a finite subset'?
 
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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