Numbers That Divide the Others' Sum
Here is problem #1841 from Mathematics Magazine,v. 83, n. 2, APRIL 2010. The problem was proposed by H. A. ShahAli, Tehran, Iran.
Let n ≥ 3 be a natural number. Prove that there exist n pairwise distinct natural numbers such that each of them divides the sum of the remaining
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Copyright © 1996-2018 Alexander BogomolnyLet n ≥ 3 be a natural number. Prove that there exist n pairwise distinct natural numbers such that each of them divides the sum of the remaining n - 1 numbers.
We need to check the claim for some small number n. Naturally,
It would have been nice if, when moving from n integers to
This indeed works: 1 divides (2 + 3 + 6), 2 divides
For a set of 5 integers we want to choose 1, 2, 3, 6, 12. Let's do the arithmetic once more:
2 divides 1 + 3 + 6 + 12,
3 divides 1 + 2 + 6 + 12,
6 divides 1 + 2 + 3 + 12,
12 divides 1 + 2 + 3 + 6.
Is there any general reason why adding the sum of the existing numbers as a new one works? Yes, there is! Let a be one of the existing numbers and A the sum of the remaining ones. So that a divides A. The new member we add equals
The answer is 1, 2, 3, 6, 12, 24, 48, ...
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