Maths Olympiad Prep

Library / /153 of 264

Number theory Difficulty 5.9 AIME, harder Prove it Romania

Let A={1,3,32,33,,32014}A = \{1, 3, 3^2, 3^3, \dots, 3^{2014}\}. We obtain a partition of AA if AA is written as a disjoint union of nonempty subsets.

a) Prove that there is no partition of AA such that the product of elements in each subset is a square.

b) Prove that there exists a partition of AA such that the sum of elements in each subset is a square.

Solution

a) Assume that such a partition exists. Then the product of all elements of AA must be a square as well. But this equals 31+2+3++2014=3201510073^{1+2+3+\dots+2014} = 3^{2015 \cdot 1007}, obviously not a square.

b) Observe that 32n+32n+1=(3n2)23^{2n} + 3^{2n+1} = (3^n \cdot 2)^2, hence a possible partition is
A={1,3}{32,33}{32012,32013}{32014}. A = \{1,3\} \cup \{3^2, 3^3\} \cup \dots \cup \{3^{2012}, 3^{2013}\} \cup \{3^{2014}\}.

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.