Math Problem

(Eugene Volokh)

Here's a math problem I much enjoyed, from the 2011 International Mathematical Olympiad:

Consider a set of four different positive integers, which we'll call a, b, c, and d. There are of course six pairwise sums of these integers, a+b, a+c, a+d, b+c, b+d, and c+d.

Some number of these pairwise sums might be divisors of the sum of all four integers, a+b+c+d. Thus, for instance, if the set is 1, 2, 4, and 8, the pairwise sums are 1+2 (3), 1+4 (5), 1+8 (9), 2+4 (6), 2+8 (10), and 4+8 (12); exactly two of these pairwise sums (3 and 5) are divisors of 1+2+4+8 (15). Or if the set if 1, 2, 4, and 10, zero of the pairwise sums are divisors of the overall sum, since that sum is 17, a prime number.

What is the largest possible number of such pairwise sums that are divisors of the overall sum? Can you have a set in which all six pairwise sums are divisors of the overall sum? Five? Four? Three? (Obviously you can have a set in which two pairwise sums are divisors of the overall sum, since that's what we see with 1, 2, 4, and 8.)

Once you identify this largest number of divisors, what are all the possible sets that yield this number? If there's an infinite number of such sets, indicate the rule through which they can be identified.

If you prefer the more rigorous formulation from the Olympiad itself, here it is: Given any set A = {a1, a2, a3, a4} of four distinct positive integers, we denote the sum a1+a2+a3+a4 by sA. Let nA denote the number of pairs (i, j) with 1 <= i < j <= 4 for which ai+aj divides sA. Find all sets A of four distinct positive integers which achieve the largest possible value of nA.