Maths Olympiad Prep

Library / Chinese Mathematical Olympiad /1 of 17

Chinese Mathematical Olympiad

Combinatorics Difficulty 4.7 AIME Prove it China

Given two integers m,nm, n greater than 11, and integers a1<a2<⋯<ama_1 < a_2 < \cdots < a_m, prove that there exists a set TT of integers with ∣T∣≤1+am−a12n+1|T| \le 1 + \frac{a_m - a_1}{2n + 1} such that each aia_i can be written as ai=t+sa_i = t + s for some t∈Tt \in T, and s∈[−n,n]s \in [-n, n]. (Posed by Leng Gangsong)

Solution

Write a1=aa_1 = a, am=ba_m = b, b−a=(2n+1)q+rb - a = (2n + 1)q + r, where q,r∈Zq, r \in \mathbb{Z} and 0≤r≤2n0 \le r \le 2n. Take

T={a+n+(2n+1)k∣k=0,1,…,q}. T = \{a + n + (2n + 1)k \mid k = 0, 1, \dots, q\}.

Then ∣T∣=q+1≤1+b−a2n+1|T| = q + 1 \le 1 + \frac{b - a}{2n + 1}. We have the set

B={t+s∣t∈T,s=−n,−n+1,…,n}={a,a+1,…,a+(2n+1)q+2n}. B = \{t + s \mid t \in T, s = -n, -n + 1, \dots, n\} = \{a, a + 1, \dots, a + (2n + 1)q + 2n\}.

Note that a+(2n+1)q+2n≥a+(2n+1)q+r=ba + (2n + 1)q + 2n \ge a + (2n + 1)q + r = b, so each aia_i belongs to BB.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.