Maths Olympiad Prep

Library / /257 of 377

Number theory Difficulty 5.3 AIME, harder Prove it United States

Problem:

Let S={s0,,sn}S=\{s_{0}, \ldots, s_{n}\} be a finite set of integers, and define S+k={s0+k,,sn+k}S+k=\{s_{0}+k, \ldots, s_{n}+k\}. We say that SS and TT are equivalent, written STS \sim T, if T=S+kT=S+k for some kk. Given a (possibly infinite) set of integers AA, we say that SS tiles AA if AA can be partitioned into subsets equivalent to SS. Such a partition is called a tiling of AA by SS.

Suppose that SS tiles the set of all integer cubes. Prove that SS has only one element.

Solution

Solution:

Let the difference between the smallest and largest element of SS be aa. Then the set equivalent to SS that contains b3b^{3} can only contain integers between b3ab^{3}-a and b3+ab^{3}+a, inclusive. But for sufficiently large bb, b3b^{3} is the only cube in this range, so SS can only have one element.

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 reproduced verbatim; metadata (topic, difficulty) added by this project.