# 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, *n* = 3

It would have been nice if, when moving from n integers to *n* + 1,*n* + 1)-set*n*-set must divide the sum of the existsing numbers. As we know little of what that sum might be, let's try the sum itself. So, let's look at the set of for numbers, 1, 2, 3,

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 *s* = *a* + *A**a* (because *A* is.) Now, for a set of *n* elements we have to establish *n* divisibility facts. For the expanded set, we need to check *n* + 1*n* hold: *a*|(*A* + *a* + *A*),*a* from the existing set of *n* integers. But the remaing *n* + 1)st*s*|*s*.

The answer is 1, 2, 3, 6, 12, 24, 48, ...

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Copyright © 1996-2018 Alexander Bogomolny