Let G be an abelian group of order k. How is the problem of minimizing the number of sums from a sequence of given length in G related to the problem of minimizing the number of k-sums? In this paper we show that the minimum number of k-sums for a sequence a 1 , . . . , a r that does not have 0 as a k-sum is attained at the sequence b 1 , . . . , b r−k+1 , 0, . . . , 0, where b 1 , . . . , b r−k+1 is chosen to minimise the number of sums without 0 being a sum. Equivalently, to minimise the number of k-sums one should repeat some value k − 1 times. This proves a conjecture of Bollobás and Leader, and extends results of Gao and of Bollobás and Leader.
IntroductionGiven a sequence a 1 , . . . , a r in Z k , the integers modulo k, a k-sum is a sum of the form a i 1 + · · · + a i k , where i 1 < · · · < i k . How large can r be without 0 being a k-sum? It is clear that we may have r = 2k − 2, by taking a 1 = · · · = a k−1 = 0 and a k = · · · = a 2k−2 = 1. Erdös, Ginzburg and Ziv [5] showed that this is best possible. In other words, they showed that if we have a 1 , . . . , a 2k−1 in Z k then some k-sum is 0. Since then, numerous other proofs of this result have been found-see Alon and Dubiner [1] for a general survey.