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Problem 1452

AIME late
Combinatorics Difficulty 5.5 Prove it HMMT November · United States · 2015

Call a set of positive integers good if there is a partition of it into two sets SS and TT, such that there do not exist three elements a,b,c∈Sa, b, c \in S such that ab=ca^{b}=c and such that there do not exist three elements a,b,c∈Ta, b, c \in T such that ab=ca^{b}=c (aa and bb need not be distinct). Find the smallest positive integer nn such that the set {2,3,4,…,n}\{2,3,4, \ldots, n\} is not good.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

First, we claim that the set {2,4,8,256,65536}\{2,4,8,256,65536\} is not good. Assume the contrary and say 2∈S2 \in S. Then since 22=42^{2}=4, we have 4∈T4 \in T. And since 44=2564^{4}=256, we have 256∈S256 \in S. Then since 2562=65536256^{2}=65536, we have 65536∈T65536 \in T. Now, note that we cannot place 88 in either SS or TT, contradiction.

Hence n≤65536n \leq 65536. And the partition S={2,3}∪{256,257,…,65535}S=\{2,3\} \cup \{256,257, \ldots, 65535\} and T={4,5,…,255}T=\{4,5, \ldots, 255\} shows that n≥65536n \geq 65536. Therefore n=65536n=65536.

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