Maths Olympiad Prep

Library / /5 of 7

, 2015

Combinatorics Difficulty 5.7 AIME, harder Prove it Romania

Given an integer n3n \ge 3 and a permutation a1,a2,,ana_1, a_2, \dots, a_n of the first nn positive integers, show that at least n\sqrt{n} distinct residue classes modulo nn occur in the list a1,a1+a2,,a1+a2++ana_1, a_1 + a_2, \dots, a_1 + a_2 + \dots + a_n.

Solution

Let rkr_k be the remainder of the partial sum a1+a2++aka_1 + a_2 + \dots + a_k upon division by nn, and let mm be the number of distinct residues in the list r1,r2,,rnr_1, r_2, \dots, r_n. Since there are mm distinct integers in the list, the number of ordered pairs with different entries that can be formed from these mm integers does not exceed m(m1)m(m-1). On the other hand, of the n1n-1 distinct ordered pairs (rk,rk+1)(r_k, r_{k+1}), k=1,2,,n1k = 1, 2, \dots, n-1, at most one has equal entries, so m(m1)(n1)1=n2m(m-1) \ge (n-1)-1 = n-2. Consequently, m(1+4n7)/2nm \ge (1+\sqrt{4n-7})/2 \ge \sqrt{n} if n4n \ge 4, the latter inequality being strict if n5n \ge 5. If n=4n=4, and a1=2a_1=2, a2=4a_2=4, a3=3a_3=3 and a4=1a_4=1, then the corresponding list of residues is 2, 2, 1, 2, so m=2=4=nm=2 = \sqrt{4} = \sqrt{n}. Finally, if n=3n=3, then every permutation produces a list consisting of exactly 2>32 > \sqrt{3} distinct residues. (If n=2n=2, the identity produces the list 1, 1, so m=1<2=nm=1 < \sqrt{2} = \sqrt{n}.)

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.